Chapter 7.7, Problem 2P

To determine

Tofind: thegreater number if smaller number is $x$ and the difference of the numbers is $8$ .

Answer to Problem 2P

The greater number is $8+x$ and the product is $p\left(x\right)=8x+x^2$ and the minimum value of product is $-16$ .

Explanation of Solution

Given information:

The difference of the numbers is $8$ and smaller number is $x$ .

Calculation: to find the greater number, let $y$ is the greater number,

Since, the difference of the numbers is $20$ ,

$\begin{array}{l}y-x=8\\ y=8+x\text{ [add]}\end{array}$

To find the product of the numbers, multiply both,

$\begin{array}{l}p\left(x\right)=x\left(8+x\right)\\ p\left(x\right)=8x+x^2\text{ [multiply]}\end{array}$

To find the maximum value of product, first differentiate and then equal to zero,

$\begin{array}{l}p\left(x\right)=8x+x^2\\ p\text{'}\left(x\right)=8+2x\text{ [differentiate]}\\ 8+2x=0\text{ [equal to zero]}\\ x=-4\text{ [simplify]}\end{array}$

Now, substitute in the product function,

$\begin{array}{l}p\left(-4\right)=8\left(-4\right)+{\left(-4\right)}^2\text{ [substitute]}\\ p\left(-4\right)=-32+16\text{ [multiply]}\\ p\left(-4\right)=-16\text{ [subtract]}\end{array}$

Thus, the minimum value of product is $-16$ .