Find two positive numbers whose sum is 8, such that the sum of the square of one number and the cube of the other number is a minimum. Use calculus to justify your solution

Answer 1

X+Y = 8

Y = 8-x

x^2 + (8-x)^3 = f(x)

f'(x)=

2x + 3(8-x)^2(-1)

2x - 3(8-x)^2 =

2x - 3(64 - 16x + x^2) =

2x - 192 + 48x - 3x^2 =

-3x^2 + 50x - 192

-3x^2 + 50x - 192 = 0

3x^2 - 50x + 192 = 0

(3x - 32)(x - 6 )=0

x=32/3 or x=6

x=32/3 = 10 and 2/3 ---> y = 8 - 10 and 2/3 = -2 and 2/3

6 and 2

Answer 2

Let x and 8 - x be the numbers..

So, we need to minimize f(x) = (8 - x)² + x³, where 0 < x < 8

f(x) = x³ + x² - 16x + 64

f'(x) = 3x² + 2x - 16 = (3x + 8)(x - 2)

f'(x) = 0 when x = -8/3 or 2

Since 0 < x < 8, the only critical number we need to consider is 2.

If 0 < x < 2, then f'(x) < 0. So, f(x) is decreasing

If 2 < x < 8, then f'(x) > 0. So, f(x) is increasing.

The minimum occurs when x = 2 and y = 6