Rectangle in a circle of radius a that maximizes x^n+ y^n

Question

Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $$x^2 + y^2 = a^2,$$ and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $$z = x^n + y^n.$$ If $n = 2$, it is clear that $z$ has the constant value $a^2$ for all rectangles.

• If $n < 2$, show that the square maximizes $z$; and
• if $n > 2$, show that $z$ is maximized by a degenerate rectangle in which $x$ or $y$ is $0$.

My approach.

$$x^2 + y^2 = a^2$$ $$\therefore\, y = (a^2 - x^2)^{1\over 2}$$

\begin{align} z &= x^n + y^n \\ {d\over dx}z & = nx^{n-1} - nx(a^2-x^2)^{n-2\over 2} \end{align}

Equating with $0$ gives: $$nx^{n-1} = nx(a^2-x^2)^{n-2\over 2}$$

$$x^{n-2} = (a^2 - x^2)^{n-2\over 2}$$

Squaring on both sides gives:

$$(x^2)^{n-2} = (a^2 - x^2)^{n-2}$$

Taking $\ln$ on both sides gives

$$(n-2) \ln(x^2) = (n-2)\ln(a^2 - x^2)$$33

Equating $x^2 = a^2 - x^2$ , we get $x = {a\over \sqrt{2}}$ and $y = {a\over\sqrt{2}}$. In particular $x=y$ implies that the optimal rectange is a square.

My computation appears to be independent of $n$, so I am puzzled about the statement in the question for $n>2$ and $n<2$.

Answer 1

${d\over dx}z$ = $nx^{n-1}$ - $nx(a^2-x^2)^{n-2\over 2}$

Equating with $0$ gives

$x= {a\over \sqrt {2}}$ so, $y = {a\over \sqrt {2}}$

and $x= 0$ so, $y = a$

Now on double differentiating z

${d^2\over dx^2}z$ = $n(n-1)x^{n-2} - n(a^2 - x^2)^{n-2\over 2} + n(n-2)x^2(a^2 - x^2)^{n-4\over 2}$

At $x= {a\over \sqrt {2}}$,

${d^2\over dx^2}z$ simplifies to be $2n(n-2)({a^2\over 2})^{n-2\over 2}$

So, if $n<2$, ${d^2\over dx^2}z$ At $x= {a\over \sqrt {2}}$ is < $0$

So, at $x= {a\over \sqrt {2}}$, Z = maximum when $n<2$ and minimum when $n>2$

When $ n < 2$ At $x =0$ ${d^2\over dx^2}z$ is undefined due to the term $n(n-1)x^{n - 2}$ = $n(n-1)$$({1\over 0})^{2 - n}$

When n>2

At $x = 0$ ${d^2\over dx^2}z$ simplifies to be $-n(a^2)^{n-2\over 2}$ which is $< 0$ as $n > 0$

So $z$ is maximum at $x = 0$

This shows that when $n<2$ maximum value of $z$ occurs when $x = y = {a\over \sqrt {2}}$ thus a square

When $n= 2$, $z$ simply equates to $a^2$

When n > 2 $z$ is maximum when $x = 0$ , $y = a$ and since $x$ and $y$ are symmetric so they are interchangeable this means z is also maximum when $x = a$ , $y = 0$ and thus it is a degenerate rectangle which gives maximum value of $z$ when $n>2$

Answer 2

Alternative way, without calculus.

• If $\,n \le 2\,$ then by the generalized means inequality $\,\left(\frac{x^n+y^n}{2}\right)^{1/n} \le \left(\frac{x^2+y^2}{2}\right)^{1/2}=\frac{a}{\sqrt{2}}\,$. The maximum is attained when the equality is achieved, for $\,x=y\,$ i.e. the square.

• If $\,n \gt 2\,$ then $\,x^n+y^n=(x^2)^{n/2}+(y^2)^{n/2} \le (x^2+y^2)^{n/2} = a^n\,$ with equality for $\,x=0\,$ or $\,y=0\,$, since $\,x^2/a^2, \,y^2/a^2 \le 1\,$ so $\,(x^2/a^2)^{n/2}+(y^2/a^2)^{n/2} \le x^2/a^2+y^2/a^2 = 1\,$.

Answer 3

For $n < 2$ your expression for $\frac{dz}{dx}$ is not defined where $y = 0$, so one must check those points separately. Also, between the two equations following, "equating it with $0$...", division by $0$ assumes that $x \neq 0$.

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Here's an alternative approach: This problem is a typical candidate for the method of Lagrange multipliers, and applying that method has the advantage that it leaves $x$ and $y$ on equal footing and in particular avoid the complications introduced by solving (partially) for $y$ in terms of $x$.

Our objective function is $$f(x, y) := x^n + y^n ,$$ and our constraint is $g(x, y) = 0$, where $$g(x, y) := x^2 + y^2 - a^2 .$$ (Herein I'll assume $a > 0$, so that the circle is nondegenerate.)

By symmetry we may as well assume that any (constrained, local) extremum $(c, d)$ lies in the closed first quadrant, i.e., are both nonnegative.

In fact, if $c = 0$, then the constraint implies $d = a$, and vice versa, so if a (constrained, local) extremum occurs where $c = 0$ or $d = 0$, its value is $f(a, 0) = f(0, a) = a^n$, and we may as well restrict the rest of our search to the interior of the first quadrant, i.e., the case $c, d > 0$.

The method of Lagrange multipliers yields that any (constrained, local) extremum $(c, d)$ satisfies $$(\nabla f)(c, d) = \lambda (\nabla g)(c, d) ,$$ that is, the system \begin{align} n c^{n - 1} &= 2 c \lambda \\ n d^{n - 1} &= 2 d \lambda . \end{align}

Dividing the first equation by the second equation and rearranging yields $c^{n - 2} = d^{n - 2}$, so either (a) $n = 2$ (which corresponds to the tautology that there is a maximum at every point on the unit circle), or (b) the only critical point is $c = d = \frac{a}{\sqrt{2}}$. In the latter case computing gives $f\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right) = a^n (\sqrt{2})^{2 - n} ,$ and whether this quantity is greater than $f(a, 0) = f(0, a) = a^n$ manifestly depends just on the sign of $2 - n$.