Question

find two positive real numbers such that they sum to 108 and the product of the first times the square of the second is a maximum

Answer 1

The two positive numbers are 36 and 72 which gives a sum equal to 108 and the product of 36 and the square of 72 is a maximum.

Any integer greater than zero is considered a positive number. A positive number can either be written as a number or with the "+" symbol in front of it.

Let us consider the two positive real numbers as x and y. Then, their sum is written as,

x+y=108

Then, y=108-x

And the product is written as,

P=xy²

Substitute value of y in the above equation, we get,

$\begin{aligned}P&=x(108-x)^2\\&=x(11664-216x+x^2)\\&=11664x-216x^2+x^3\\&=x^3-216x^2+11664x\end{aligned}$

Now, differentiate P with respect to x, and we get,

$\begin{aligned}\frac{dP}{dx}&=3x^2-432x+11664\\&=x^2-144x+3888\end{aligned}$

Solving the above equation to zero, we get,

$\begin{aligned}x^2-144x+3888&=0\\x^2-36x-108x+3888&=0\\x(x-36)-108(x-36)&=0\\(x-108)(x-36)&=0\\x&=\text{108 or 36}\end{aligned}$

Substitute values of x in y=108-x, to get values of y.

If we substitute x=108, we get the y value as zero which doesn't give the required solution.

But, if we substitute x=36, we get,

$\begin{aligned}y&=108-36\\y&=72\end{aligned}$

Thus, the two positive numbers are 36 and 72.

To know more about positive numbers:

brainly.com/question/1635103

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