# Chaitanya's Random Pages

## November 28, 2010

### Ashes Preview

Filed under: sport — ckrao @ 9:10 pm

The Ashes is one of cricket’s main contests, held twice every four years between the two oldest teams England and Australia. The buildup to this particular test series has been as exciting as I have seen. England appear to have their best chance of winning in Australia since they last did so in the 1986-7 season. They have prepared very well and most of their key players appear to be in good form. Amazingly, out of 65 series played between the two teams, only 5 of them have been drawn series – of the remaining 60 Australia narrowly leads the head-to-head 31-29 [1].

Australia recently had one of the most dominant periods for any team. Between early 1999 and the end of 2007 they only lost 2 matches at home! More about this period of dominance here and here. A key to this was being blessed with so many great players born within a short period. Below is a list of the number of Australian test cricketers born in each year of the 1970s.

 Year of birth Number of Australian Test players 1970 9 1971 8 1972 4 1973 0 1974 5 1975 5 1976 1 1977 3 1978 0 1979 4

Note how many players were born from 1970-1971. It is quality of this group that is mind-blowing and is what largely prevented many players born in the late 1970s to represent their country. Consider this list:

born in 1970: McGrath (124 tests, 563 wickets @ 21.64), Langer (105), Slater (74), Lehmann (27, averaged 57+ in first class cricket), Bevan (18, one of the best ODI players, also averaged 57+ in first class cricket), Fleming (20, 75 wickets @ 25.89), Bichel (19, many 12th man appearances)

born in 1971: Gilchrist (96 tests, 416 dismissals, one of the most destructive batsmen I have ever watched), Hayden (103, 8625 runs @ 50.74), Martyn (67), MacGill (44, 208 wickets @ 29.02), Blewett (46), Elliot (21, also averaged 47 in first class cricket)

Add to this stalwarts in Warne, the Waugh twins, Ponting as well as firepower bowling in Lee and Gillespie, and it was a formidable combination!

Now all of those players but Ricky Ponting have retired and Australia’s test ranking now sits at 5th. Their home record is no longer at the dizzying heights of 1999-2007, having lost a series in 2008-9 for the first time in 20 years. At the Gabba (Brisbane) their record remains formidable with no losses since the 1988-9 series. England would be happy to leave that match with a draw. There is an interesting article about the varying fortunes of visiting players in Adelaide and Sydney versus the remaining grounds here.

Let’s see how the teams for the first test stack up in terms of their batting/bowling ratings (according to cricketratings.com). Note that a rating of 700 is considered very good. Interestingly the highest ranked batsman from either team only has a current rank of 14 (Australian vice captain Michael Clarke). It would appear that England’s strength is in its bowling, boasting 3 players in the top 11 and 2 in the top 5. Australia’s most successful bowlers in the post Warne-McGrath era have been Stuart Clark and Mitchell Johnson but the latter has not fired in recent times. I expect Bollinger to play a part in the series too.

Test cricket ratings (world rankings in parentheses)

 AUSTRALIA ENGLAND Batting Bowling Batting Bowling Watson 647 (25) 441 (35) Strauss 644 (26) Katich 735 (15) 127 (75) Cook 628 (30) Ponting 731 (17) Trott 733 (16) Clarke 738 (14) 113 (77) Pietersen 648 (24) Hussey 626 (31) Collingwood 617 (32) 90 (83) North 511 (45) 202 (62) Bell 643 (27) Haddin 553 (40) Prior 611 (34) Johnson 317 (89) 735 (7) Broad 490 (49) 650 (11) Doherty DEBUT Swann 335 (82) 858 (2) Siddle 562 (19) Finn 484 (29) Hilfenhaus 560 (20) Anderson 744 (5)

## November 20, 2010

### A collection of algebraic identities

Filed under: mathematics — ckrao @ 5:35 am

In this post I will mainly focus on proving the following nice result that a friend gave to me. Then I will list some other nice identities I have collected from books and the internet.

Let $z_1, z_2, \ldots, z_n$ be n distinct numbers (they could even be complex-valued). Then

$\displaystyle \sum_{i=1}^n \prod_{j\neq i} \frac{1}{z_i-z_j} = 0.\quad \quad ... (1)$

For example, for the case n = 3 we have

$\displaystyle \frac{1}{(z_1 - z_2)(z_1 - z_3)} + \frac{1}{(z_2 - z_3)(z_2 - z_1)} + \frac{1}{(z_3 - z_1)(z_3 - z_2)} = 0.$

To prove (1), consider the following partial fraction expansion:

$\displaystyle \frac{1}{\prod_{i=1}^n (z-z_i)} = \sum_{i=1}^n \frac{A_i}{z-z_i}.\quad \quad ... (2)$

To find the $A_i$, we multiply both sides of (2) by $(z-z_i)$ and then set $z$ to $z_i$, for each $i$. All terms on the right become zero except the one with $A_i$:

$\displaystyle \prod_{j\neq i} \frac{1}{z_i-z_j} = A_i.\quad \quad ... (3)$

Next, multiply both sides of (2) by $\prod_{i=1}^n (z-z_i)$ to obtain

$\displaystyle 1= \sum_{i=1}^n A_i\prod_{j \neq i} (z-z_i).\quad \quad ... (4)$

Taking the coefficient of $z^{n-1}$ of both sides of (4) (treated as a polynomial in z) gives

$\displaystyle 0 = \sum_{i=1}^n A_i,$

which after comparing (1) and (3) is what we wished to show. Note that taking coefficients of other powers of $z$ in both sides of (4) reveals more identities.

I found this result quite cute and it made me dig up other identities I have seen over the years. I will prove one more and the rest can be left as exercises for the interested reader.

If $x + y + z = 0$,

$\displaystyle \left(\frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y}\right)\left(\frac{z}{x-y} + \frac{x}{y-z} + \frac{y}{z-x}\right) = 9.$

An easy way of doing this eludes me. Adding the first and second terms inside the first pair of brackets gives

$\begin{array}{lcl} \frac{x-y}{z} + \frac{y-z}{x} &=& \frac{x^2 - xy + zy - z^2}{zx}\\&=& \frac{(x-z)(x+z-y)}{zx}\\&=&\frac{(x-z)(-2y)}{zx},\end{array}$

where in the last equality we used the fact that $x + y + z = 0$. Multiplying this by $\frac{y}{z-x}$ in the right bracket gives

$\begin{array}{lcl} \frac{-2y^2(x-z)}{(z-x)zx} &=& \frac{2y^2}{zx}.\end{array}$

Summing this cyclically over the other possible products gives

$\begin{array} {lcl} \left(\frac{x-y}{z} + \frac{y-z}{x} + \frac{z-x}{y}\right)\left(\frac{z}{x-y} + \frac{x}{y-z} + \frac{y}{z-x}\right) &=& 1 + 1 + 1 + \frac{2y^2}{zx} + \frac{2z^2}{xy} + \frac{2x^2}{yz}\\&=& 3 + 2 \frac{x^3 + y^3 + z^3}{xyz}\\& = & 3 + 2.3\\ &=& 9,\end{array}$

where the penultimate equality follows from one of the identities shown below.

Finally here is a list of interesting identities, with potential to grow. We assume they are defined for complex numbers for which any denominator shown is non-zero.

### Algebraic Identities

• $\displaystyle x^2 - y^2 = (x+y)(x-y)$ (elementary but very useful!)
• More generally, $\displaystyle x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$
• If $n$ is odd, $\displaystyle x^n + y^n = (x+y)(x^{n-1} - x^{n-2}y + \ldots - xy^{n-2} + y^{n-1}).$
• Binomial theorem: $\displaystyle (x+y)^n = \sum_{k=0}^n \binom{n}{k}x^ky^{n-k}$
• In particular, $\displaystyle (x+y)^2 = x^2 + 2xy + y^2, \ (x+y)^3 = x^3 + 3xy(x+y) + y^3.$
• $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$
• Multinomial theorem: $\displaystyle \left(x_1 + x_2 + \ldots + x_m\right)^n = \sum_{k_1+k_2 + \ldots k_m=n}\binom{n}{k_1, k_2, \ldots, k_m}x_1^{k_1}x_2^{k_2}\ldots x_m^{k_m},$
where $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m} = \frac{n!}{k_1! k_2! \ldots k_m!}.$
• $\begin{array}{lcl} x^3 + y^3 + z^3 - 3xyz &=& (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)\\ &=& \frac{1}{2}(x + y + z)\left((x-y)^2 + (y-z)^2 + (z-x)^2\right) \end{array}$
(this implies $x^3 + y^3 + z^3 = 3xyz$ if $x + y + z = 0)$
• $\displaystyle (x-y)^3 + (y-z)^3 + (z-x)^3 = 3(x-y)(y-z)(z-x)$ (follows from the previous result)
• $\displaystyle x^4 + 4y^4 = x^4 + 4x^2y^2 + 4y^4 - 4x^2 y^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$
• $\displaystyle x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$
• $\displaystyle \left(\sum_i x_i^2 \right)\left(\sum_i y_i^2 \right) = \left(\sum_i x_i y_i\right)^2 + \sum_{i < j} (x_i y_j - x_j y_i)^2$ (Lagrange’s identity)
Complex case: $\displaystyle \left(\sum_i |x_i|^2 \right)\left(\sum_i |y_i|^2 \right) = \left|\sum_i x_i y_i\right|^2 + \sum_{i < j} \left|\overline{x_i} y_j - x_j \overline{y_i}\right|^2$
• Special case: $\displaystyle (a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad + bc)^2 = (ac + bd)^2 +(ad - bc)^2.$ (Brahmagupta-Fibonacci identity, or complex number multiplication)
• $\displaystyle (a^2 + nb^2)(c^2 + nd^2) = (ac - nbd)^2 + n(ad+bc)^2 = (ac + nbd)^2 + n(ad - bc)^2.$ (proving that numbers of the form $x^2 +ny^2$ are closed under multiplication – see the Wikipedia entry on the Brahmagupta-Fibonacci identity)

$\begin{array}{l}(x_1^2 + x_2^2 + x_3^2 + x_4^2)(y_1^2 + y_2^2 + y_3^2 + y_4^2) \ = \\ \quad (x_1y_1 - x_2y_2 - x_3y_3 - x_4y_4)^2 \ + \\ \quad (x_1y_2 + x_2y_1 + x_3y_4 - x_4y_3)^2 \ + \\ \quad (x_1y_3 - x_2y_4 +x_3y_1 + x_4y_2)^2 \ + \\ \quad (x_1y_4 + x_2y_3 - x_3y_2 + x_4y_1)^2\end{array}$

• $\displaystyle (x + y + z)^3 - (x^3 + y^3 + z^3) = 3(x+y)(y+z)(z+x)$
• $\displaystyle (x + y + z)^5 - (x^5 + y^5 + z^5) = 5(x+y)(y+z)(z+x)(x^2 + y^2 + z^2 + xy + yz + zx)$
• $\displaystyle (x + y + z)(xy + yz + zx) - xyz = (x + y)(y+z)(z+x)$
• $\displaystyle x^2(y+z) + y^2(z+x) + z^2(x+y) + 2xyz = (x + y)(y+z)(z+x)$
• $\displaystyle xyz(x^3 + y^3 + z^3) - (x^3y^3 + y^3z^3 + z^3 x^3) = (x^2-yz)(y^2 - zx)(z^2 - xy)$
• $\displaystyle (x^2 + y^2 + z^2)^2 - 2(x^4 + y^4 + z^4) = (x + y + z)(-x + y + z)(x - y + z)(x + y -z)$
(used in Heron’s formula for the area of a triangle given side lengths $x, y, z$)
• If $x + y + z = xyz$, $\displaystyle x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2) = 4xyz.$
• $\displaystyle \sum_{\text{cyc}}(zx + 1)(yz+1)(x-y) = \sum_{\text{cyc}}x^2(z-y)$
• $(x+y)^5 - x^5- y^5 = 5xy(x+y)(x^2 + xy + y^2)$
• $(x+y)^7 - x^7 - y^7 = 7xy(x+y)(x^2 + xy + y^2)^2$
• $(x+y)^9 - x^9 - y^9 = 3xy(x+y)\left[3(x^2 + xy + y^2)^3 + x^2y^2(x+y)^2\right]$
• $\displaystyle \sum_{\text{cyc}} \frac{x^3}{(x-y)(x-z)} = x + y + z$
• $\displaystyle \sum_{\text{cyc}} \frac{x-y}{z} + \prod_{\text{cyc}} \frac{x-y}{z} = 0$

Here the $\displaystyle \sum_{\text{cyc}}$ or $\displaystyle \prod_{\text{cyc}}$ notation is used to denote a cyclic sum or product through the indices $x,y,z$, so that the three permutations (x,y,z), (y,z,x) and (z,x,y) are used. It is also worth knowing that any symmetric polynomial can be written as a polynomial of elementary symmetric polynomials (which are sums of all possible products of a fixed number of the variables, for example $x + y + z, xy + yz + zx$, and $xyz$).

## November 14, 2010

### End of a remarkable heat streak

Filed under: climate and weather,geography — ckrao @ 5:42 am

I have been tracking the maximum temperatures of Makkah, Saudi Arabia for the last few weeks and noticed that even this late in the year it has been recording temperatures in excess of 100°F. The Guinness Book of Records has listed Marble Bar, Western Australia as the place with the most consecutive days above 100°F with 160, which it set back in the summer of 1923-24. It would appear that in recent times Makkah exceeds this almost every year. Makkah (or Mecca) is of course famous as the holiest place for Islam (the birthplace of Muhammad), with millions going there on pilgrimage for Hajj every year.

From March 31 to November 12 this year, if the weather records are reliable, Makkah has reported a maximum temperature in excess of 100°F every single day – 227 consecutive days!! The maximum on November 13 was a chilly 99.3°F (37.4°C). 🙂

This comfortably breaks its previous record of 208 days in 2007 (April 17 – November 10). I have graphed the maximum temperatures from March 1 to November 20 December 8 2010 here. Even as late as December 2 the temperature reached 38°C!

If the numbers are right the amazing feature of this streak is the consistency of temperatures. Within the 227-day streak above the pink line, 199 days were at least 40°C but only 42 of these were at least 45°C. Other parts of the middle east are warmer in the middle of summer. It is only in October and November where it becomes one of the hottest few places on the planet. If it were not for a single day in the middle of July the streak of 40+°C days would also be huge – something like 162 days! It appears that 100°F is a perfectly chosen threshold for such a heat streak.

After seeing this, I tried to find the place that had the hottest streak of 45+°C temperatures this year. The best I could find was a place in the Kuwaiti desert called Jahra (another station Mitribah showed hotter temperatures but apparently the measurements there are not so trustworthy). Here is a graph of the maximum temperatures of Jahra also dating back to 1 March of this year.

Between 1 July and 10 September (inclusive), its lowest maximum temperature was 45.3°C – that’s 72 consecutive days above 45°C! It could have been many more were it not for a 5-day “cool” spell at the end of June. You can see how high the temperatures were between June and September, and then they fell away in October while Makkah’s heat remained.

Finally, you might be wondering how this level of heat compares with another hot spot, Death Valley in California. This year’s summer was not as intense as that of 2009, so I will plot 2009 temperatures here (starting a month later than the previous plots and finishing earlier):

The vertical red lines mark July 1 and July 31 to demarcate one of the hottest months (in maximum temperature) recorded anywhere in the world in the past 50 years. The average maximum temperature in July 2009 for Death Valley was 49.2°C (120.6°F)! For comparison in Jahra July and August this year had 48.1°C and 48.2°C as the average maximum temperatures respectively. In Death Valley the longest streak of 45+ °C days was 29 – far short of the 72 recorded at Jahra this year (if measurements there are reliable). Some places in the Middle East have longer heat streaks, but during the middle of summer, it is hard to match Death Valley for peak heat.

References:

http://www.ogimet.com (Synop report summaries for Makkah and Jahra)

http://accuweather.com (for Death Valley data)

## November 7, 2010

### The Ladder around a Corner Problem

Filed under: mathematics — ckrao @ 9:21 am

This post is inspired by Question 19 of this year’s University Maths Olympics (at the University of Melbourne):

Dana is moving into a mansion, and she needs a huge whiteboard. This whiteboard is to be
so heavy that it cannot be lifted; only wheeled. A certain corridor is 2.7 metres wide, and
is perpendicular to a 6.4 metre wide corridor. What is the length (in metres) of the longest
whiteboard that she can manoeuvre around the corner?

I had never seen this type of problem and was intrigued by it. If you like have a go at it yourself before reading further. I will first show my initial approach and then a nice solution that I looked up.

The first observation is that the longest ladder would touch the two walls and the corner of the inside of the corridor as illustrated above. I will define x and y as in the diagram below:

By similar triangles,

$\displaystyle \frac{y-2.7}{6.4} = \frac{2.7}{x-6.4}, \quad ...(*)$

from which

$\displaystyle y = \frac{2.7 \times 6.4}{x-6.4} + 2.7$

This can be viewed as one branch of a hyperbola, illustrated in green here (using Graph v4.3):

I have also plotted in red a circular arc of radius 12, corresponding to the possible (x,y) locations of a whiteboard of length 12. The longest such whiteboard should be tangential to the hyperbola, so we have the following maximisation problem:

$\displaystyle \max x^2 + y^2 \quad \text{subject to} \quad y \leq \frac{2.7 \times 6.4}{x-6.4} + 2.7, \quad x > 6.4.$

Then we write

$\begin{array}{lcl} x^2 + y^2 &\leq& x^2 + \left(\frac{2.7 \times 6.4}{x-6.4} + 2.7\right)^2\\&=& x^2 + 2.7^2\left(\frac{6.4}{x-6.4} + 1\right)^2.\end{array}$

We wish to minimise the right side viewed as a function of x. Denoting this quantity by z and setting $dz/dx = 0$ gives

$\displaystyle \begin{array}{lcl}0 &=& 2x + 2.7^2\times 2\left(\frac{6.4}{x-6.4}+1\right)\left(\frac{-6.4}{(x-6.4)^2}\right)\\&=&2x-2.7^2\times 2 \times 6.4 \frac{x}{(x-6.4)^3}.\end{array}$

Dividing both sides by $2x$ and rearranging, we obtain

$\displaystyle (x-6.4)^3 = 2.7^2 \times 6.4 = \left(\frac{3^3}{10}\right)^2 \times \frac{4^3}{10} = \left(\frac{3^2 \times 4}{10}\right)^3 .$

Hence $x -6.4 = \frac{36}{10} = 3.6$, from which $y = \frac{2.7 \times 6.4}{x-6.4} + 2.7 = \frac{2.7 \times 6.4}{3.6} + 2.7 = 4.8 + 2.7 = 7.5$ and the maximum whiteboard length is

$\displaystyle \sqrt{x^2 + y^2} = \sqrt{10^2 + 7.5^2} = 2.5\sqrt{4^2 + 3^2} = 2.5 \times 5 = 12.5 \text{ metres}$.

[This value of x indeed corresponds to the minimum as against a maximum, since dz/dx is increasing for x > 6.4: it increases from $-\infty$ and behaves like 2x for large x.]

So what is the relationship between the answer 12.5 and the corridor widths 6.4 and 2.7? Redoing the above calculation with 6.4 and 2.7 replaced by a and b gives us $\displaystyle x = a + a^{1/3}b^{2/3}, y = b + a^{2/3}b^{1/3}$ and the following remarkable formula for the maximum whiteboard length:

$\displaystyle (a^{2/3} + b^{2/3})^{3/2}$

Looking it up, this is seen as a “ladder around a corner” problem often used in calculus text books (e.g. see [1], [2]). The above form suggests a proof using non-calculus means. The similar triangle relationship $\displaystyle (x-a)/b = a/(y-b)$ in (*) can be rewritten as

$\displaystyle \frac{a}{x} + \frac{b}{y} = 1.$

We wish to find the shortest distance between the origin and the hyperbola $\displaystyle a/x + b/y = 1$. This reminds me of the better known problem of finding the distance between a point and the straight line $x/a + y/b = 1$.  This is found to be $\displaystyle ab/\sqrt{a^2 + b^2}$ via the Cauchy-Schwarz inequality as follows. This inequality can be remembered as the dot product of two vectors being less than or equal to the product of their lengths:

$\displaystyle 1 = \frac{x}{a} + \frac{y}{b} \leq (x^2 + y^2)^{1/2}\left(\frac{1}{a^2} + \frac{1}{b^2}\right)^{1/2} = (x^2 + y^2)^{1/2} \frac{\sqrt{a^2 + b^2}}{ab}$

Rearranging gives $\displaystyle (x^2 + y^2)^{1/2} \geq \frac{ab}{a^2 + b^2}.$ Another easy way to see this is by calculating the area of a right triangle with legs having length a and b in two ways ($ab = h \sqrt{a^2 + b^2}$), giving us the formula for h, the distance from the origin to the hypotenuse).

The generalisation of the Cauchy-Schwarz inequality is Hölder’s inequality, which replaces the 2-norm with p- and q-norms where $p + q = pq$ (or $q = p/(p-1)$):

$\displaystyle \left(\sum_i |u_i|^p\right)^{1/p}\left(\sum_i |v_i|^q\right)^{1/q} \geq \sum_i |u_i v_i|$

(This in turn can be proved via the AM-GM inequality, which I may elaborate on in a future post. Equality holds if and only if for all i, $|u_i|^p = k|v_i|^q$ for some constant $k$. The inequality reduces to the Cauchy-Schwarz inequality in the case  p = q = 2.)

We use this inequality in our case with $p = 3, q = 3/2$ as follows:

$\begin{array}{lcl} x^2 + y^2 &=& (x^2 + y^2)(\frac{a}{x} + \frac{b}{y})^2\\&=&\left\{\left(\left(x^{2/3}\right)^3 + \left(y^{2/3}\right)^3\right)^{1/3}\left(\left(\frac{a^{2/3}}{x^{2/3}}\right)^{3/2} + \left(\frac{b^{2/3}}{y^{2/3}}\right)^{3/2}\right)^{2/3}\right\}^3\\&\geq& \left(a^{2/3} + b^{2/3}\right)^3\end{array}$

The idea used above is to manipulate the right side so that the $x$ and $y$ terms are able to cancel upon application of the inequality. Equality holds if and only if $x^2 = ka/x$ and $y^2 = kb/y$, or in other words, $\frac{x}{a^{1/3}} = \frac{y}{b^{1/3}}$, which when combined with the condition $\displaystyle 1 = \frac{x}{a} + \frac{y}{b}$ implies $\displaystyle x = a + a^{1/3}b^{2/3}, y = b + a^{2/3}b^{1/3}$ as found earlier.

### Generalisation

The above method can be readily generalised. Suppose we wish to minimise $x^{\alpha} + y^{\alpha}$ given $ax^{\beta} + by^{\beta} = 1$ (where $\alpha > 0 > \beta$ and $a,b >0$). If we redo the calculation above, for the cancellation in x and y to occur we need to choose p and q such that

$\displaystyle x^{\alpha/p + \beta/q} = x^0$.

Since $q = p/(p-1)$ this means $\alpha/p + \beta(p-1)/p = 0$, or $p = \frac{\beta - \alpha}{\beta}, q = \frac{\alpha - \beta}{\alpha}$. We then are able to write

$\begin{array}{lcl} x^{\alpha} + y^{\alpha} &=& (x^{\alpha} + y^{\alpha})(ax^{\beta} + by^{\beta})^{-\frac{\alpha}{\beta}}\\&=&\left(\left(x^{\alpha/p}\right)^p + \left(y^{\alpha/p}\right)^p\right)\left(\left(\left[ax^{\beta}\right]^{\frac{1}{q}}\right)^q + \left(\left[ax^{\beta}\right]^{\frac{1}{q}}\right)^q\right)^{-\frac{\alpha}{\beta}}\\&=&\left\{\left(\left(x^{\frac{\alpha \beta}{\beta - \alpha}}\right)^{\frac{\beta-\alpha}{\beta}} + \left(y^{\frac{\alpha \beta}{\beta - \alpha}}\right)^{\frac{\beta-\alpha}{\beta}}\right)^{\frac{\beta}{\beta-\alpha}}\left(\left(\left[ax^{\beta}\right]^{\frac{\alpha}{\alpha-\beta}}\right)^{\frac{\alpha-\beta}{\alpha}} + \left(\left[by^{\beta}\right]^{\frac{\alpha}{\alpha-\beta}}\right)^{\frac{\alpha-\beta}{\alpha}}\right)^{\frac{\alpha}{\alpha-\beta}}\right\}^{\frac{\beta-\alpha}{\beta}}\\&\geq& \left(a^{1/q} + b^{1/q}\right)^p\\&=&\left(a^{\alpha/(\alpha-\beta)} + b^{\alpha/(\alpha-\beta)}\right)^{\frac{\beta-\alpha}{\beta}}.\end{array}$

The minimum value is attained when $x^{\alpha} = kax^{\beta}$ and $y^{\alpha} = kby^{\beta}$, or $\displaystyle \frac{x}{a^{1/(\alpha- \beta)}} = \frac{y}{b^{1/(\alpha- \beta)}}$. This implies

$\displaystyle x = \left[a + \left(\frac{b}{a}\right)^{\frac{\beta}{\alpha - \beta}}\right]^{-\frac{1}{\beta}}, \quad y = \left[b + \left(\frac{a}{b}\right)^{\frac{\beta}{\alpha - \beta}}\right]^{-\frac{1}{\beta}}.$

### Lagrange multipliers

A useful technique in calculus to attack such problems is via Lagrange multipliers. The principle there is that at the optimum, the gradients of the objective and constraint function are parallel. In the above graph, imagine where tangents of the green and red curve match. This gives us the condition

$\displaystyle \left(\alpha x^{\alpha-1}, \alpha y^{\alpha-1}\right) = \lambda \left(a\beta x^{\beta-1}, b\beta y^{\beta-1} \right),$

leading to $\displaystyle \frac{x}{a^{1/(\alpha- \beta)}} = \frac{y}{b^{1/(\alpha- \beta)}}$ as before.

### Final Notes

1. There is also a useful physical interpretation of the original problem mentioned in [3]. Consider the forces applied to the ladder by the corner and two walls. In equilibrium the sum of these forces and torques applied to the ladder is zero. This gives a set of equations that leads to the above solution.

2. If instead of a one-dimensional segment (whiteboard/ladder) we considered a two dimensional rectangle (e.g. a couch) the problem apparently becomes far more involved, requiring the solution of a sixth degree polynomial. See [4] for more details.

### References

[1] Atsina, C. & Nelson, R., When Less is More: Visualising Basic Inequalities, 2009, p132.

[2] Wrede, R. and Siegel, M., Advanced Calculus, 2002, p85.

[3] Levi, M., The Mathematical Mechanic, 2009, p65.