Chapter 4.4, Problem 53E

To determine

To identify: The correct option.

Answer to Problem 53E

D

Explanation of Solution

Given information: The sum of a two positive numbers is 60.

Consider that the two numbers are $x$ and $y$ .

So, according to the given statement, $x+y=60$ or $y=60-x$ .

It is given that one number is equal to the square of a second number.

So, $P=x^{2}y$ , where $P$ is the product is the product of $x^2$ and $y$ .

Substitute $y=60-x$ into the equation, $P=x^{2}y$ , and simplify it.

  $\begin{array}{c}P=x^2\left(60-x\right)\\ =60x^2-x^3\end{array}$

Differentiate the obtained equation with respect to $x$ .

  $\frac{dP}{dx}=120x-3x^2$

Equate the obtained derivative to 0 and solve for $x$ .

  $\begin{array}{c}\frac{dP}{dx}=0\\ 120x-3x^2=0\\ x=0,40\end{array}$

If $x=0$ , then the product $P$ would also be 0.

So, the value of $x$ is only 40.

Substitute $x=40$ into the equation, $y=60-x$ , and solve for $y$ .

  $\begin{array}{c}y=60-40\\ =20\end{array}$

Now, find the product $P$ by substituting $x=40$ and $y=20$ into the equation, $P=x^{2}y$ , and simplify it to determine the required product.

  $\begin{array}{c}P={\left(40\right)}^2\left(20\right)\\ =32000\end{array}$

As a result, option (D) is correct.