# Among all pairs of numbers whose sum is 16, find a pair whose product is as large as possible....

## Question:

Among all pairs of numbers whose sum is 16, find a pair whose product is as large as possible. What is the maximum product?

## Maxima or Minima:

To evaluate the maximum or minimum point of any function {eq}y=f(x) {/eq}, first, we calculate the critical points by evaluating the derivative of the function and then set the derivative equal to zero. Then we'll apply the second derivative test to check maxima or minima.

• Maxima occur when the second derivative is less than zero at the critical point.
• Minima occurs when the second derivative is greater than zero at the critical point.

Let the two numbers be {eq}x \ and \ y {/eq}.

According to the given condition {eq}x+y=16 {/eq} for which {eq}xy {/eq} is maximum.

From the given equation:

{eq}x+y=16 \Rightarrow y=16-x {/eq}

Therefore

{eq}\begin{align} f(x) &=xy\\ &=x(16-x)\\ &=16x-x^2 \end{align} {/eq}

A maximum occurs when the second derivative is less than zero at the critical points.

First, we calculate the critical points.

To evaluate critical points, we first calculate the first derivative and then set it equal to zero.

{eq}\begin{align} f'(x) &=\dfrac{d}{dx}(16x-x^2)\\ &=16(1)-2x\\ &=16-2x \end{align} {/eq}

Set this derivative equal to zero.

{eq}\begin{align} f'(x) &=0\\ 16-2x &=0\\ 16 &=2x\\ x &=8 \end{align} {/eq}

Now, we calculate the second derivative of the function {eq}f(x) {/eq} to check maxima or minima.

{eq}\begin{align} f''(x) &=\dfrac{d}{dx}(16-2x)\\ &=0-2(1)\\ &=-2 \end{align} {/eq}

At {eq}x=8 {/eq}, {eq}f''(x)=-2<0 {/eq}

Therefore, the maximum of product occurs at {eq}x=8 {/eq}.

For the value of {eq}x=8 {/eq}, the value of {eq}y {/eq} is

{eq}\begin{align} y &=16-x\\ &=16-8\\ &=8 \end{align} {/eq}

Therefore, the pair of numbers whose sum is 16 and product is maximum is {eq}\color{blue}{\boxed{(8,8)}} {/eq}

Maximum product={eq}\color{blue}{\boxed{xy=(8)(8)=64}} {/eq} 