Find two positive numbers such that the sum of the one and the square of the other is 200 and whose product is a maximum

Answer 1

Let the number to be squared be x and the other number be y; then, x^2 + y = 200, or y = 200 - x^2

We are trying to maximize xy=x(200-x^2) = 200x-x^3.

We see why the values had to be positive to make sense, as the function goes to infinity as x goes to minus infinity.

Note that 200x-x^3 = -x(x^2-200)=-x(x-10√2)(x+10√2)

We can see that the maximum for x > 0 will be on (0,10√2), as that is where the function is positive.

There will be a local minimum on (-10√2,0)

Take the derivative of f(x) = 200x-x^3

f'(x) = 200 - 3x^2

f'(x) = 0 at x = ±√(200/3)

The local maximum is at √(200/3) or 10√6/3

y = 200 - x^2 = 200 - 200/3 = 400/3 or 133 1/3

The maximum is at (10√6/3, 133 1/3)