High School How to Maximize the Product ab Given Specific Square Root Conditions?

[anemone]

Here is this week's POTW:

-----

Let $a,\,b$ be non-negative numbers with $$\sqrt{1-\frac{a^2}{4}}+\sqrt{1-\frac{b^2}{16}}=\frac{3}{2}$$.

Find the maximum value of $ab$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. castor28
2. kaliprasad

Solution from castor28:

Spoiler

Let us write $x=\dfrac{a^2}{4}$, $y=\dfrac{b^2}{16}$. Maximizing $xy$ is equivalent to maximizing $ab$, but the equations are simpler.

The domain of interest is an arc of parabola between the points $\left(0,\dfrac34\right)$ and $\left(\dfrac34,0\right)$; in particular, we have:

$$ 0 \le x,y\le\frac34$$

The problem can be restated as: maximize $xy$ subject to the condition $\sqrt{1-x} + \sqrt{1-y} = \dfrac32$.

We define the Lagrange function:

$$L = xy +\lambda\left(\sqrt{1-x} + \sqrt{1-y} - \dfrac32\right)$$
​

and compute the partial derivatives:

$$\begin{align*}
\frac{\partial L}{\partial x} &= y - \frac{\lambda}{2\sqrt{1-x}} = 0\\
\frac{\partial L}{\partial y} &= x - \frac{\lambda}{2\sqrt{1-y}} = 0
\end{align*}$$
​

Eliminating $\lambda$, we get:

$$\begin{align*}
&\lambda = 2y\sqrt{1-x}= 2x\sqrt{1-y}\\
&y^2(1-x) = x^2(1-y)\\
&(x-y)(x+y-xy) = 0
\end{align*}$$
​

Because $0\le x,y<1$, we have $(x+y-xy)> 0$ (unless $x=y=0$, which does not satisfy the equation). This shows that there is only one extremum inside the domain. As we have $xy=0$ on the endpoints and $xy>0$ inside, this is indeed a maximum.

Note that it is a priori obvious that there is an extremum for $x=y$, because of the symmetry of the expression; however, we had to prove that this is the only extremum.

With $x=y$, the equation becomes:

$$2\sqrt{1-x}=\dfrac32$$
​

giving $x=y=\dfrac{7}{16}$, $a = \dfrac{\sqrt7}{2}$, $b=\sqrt7$, and $ab=\dfrac72$.

Alternative solution:

Spoiler

By using trigonometric substitution, we let

$\dfrac{a}{2}=\sin x$ and $\dfrac{b}{4}=\sin y$

Therefore the condition becomes $\cos x+\cos y=\dfrac{3}{2}$ and we need to maximize $\sin x \sin y$.

Apply the AM-GM inequality we get

$\begin{align*}\sin x \sin y &\le \dfrac{\sin^2 x+\sin^2 y+2\sin x \sin y}{4}\\&=\dfrac{2\sin x \sin y+2-\left(\dfrac{9}{4}-2\cos x \cos y \right)}{4}\\&=\dfrac{2\cos (x-y)-\dfrac{1}{4}}{4}\\& \le \dfrac{\dfrac{7}{4}}{4}\\&=\dfrac{7}{16}\end{align*}$

Hence, the maximum of $ab$ is $\dfrac{7}{2}$.