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Find two positive real numbers x and y such that they add to 120 and y(x^2) is as large as possible.

User Huske
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1 Answer

1 vote

Answer:

(x, y) = (80, 40)

Explanation:

Since the x-factor is squared, its relative contribution to the sum will be twice that of the y-factor. The two numbers are (x, y) = (80, 40).

_____

There are several ways to work this problem. One is to express y in terms of x, then maximize the resulting function of x.

y = 120 -x

f(x,y) = y(x^2)

f(x) = (120 -x)(x^2) = -x^3 +120x^2

Then the derivative of the product with respect to x is this.

f'(x) = -3x^2 +240x

We want to set that equal to zero and find the corresponding value of x.

-3x(x -80) = 0

x = 0 or 80

Since the leading coefficient of the cubic is negative, we know the smallest x-value will correspond to a minimum. We want the maximum, so we want ...

x = 80

y = 120 -80 = 40

The two real numbers of interest are (x, y) = (80, 40).

Find two positive real numbers x and y such that they add to 120 and y(x^2) is as-example-1
User Donovan Thomson
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