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

  (x, y) = (80, 40)

Step-by-step 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).

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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).

if he has no friends he cant give them out to his friends

step-by-step explanation:

no friends = cant give anything out

i would say c is the best answer, hope it's right!

good luck! =)

cutie~