Problem involving derivatives

Sum of two positive numbers are 20. Find those numbers if the sum of one number squared and other number cubed is lowest possible.

Let x be one number. The other number is 20-x

So you want to minimize (20-x)^2 + x^3

f(x) = x^3 + (20-x)^2

f'(x) = 3x^2 -2(20-x)=0

3x^2 +2x - 40=0
(x+4)(3x-10)=0
x= -4 or x= 10/3

x must be positive so x= 10/3
The two numbers are 10/3 and 50/3

Hi,
we only need to find one number (x), the other one will automatically be 20-x
So we need to find the minimum of the function

f(x) = x² + (20-x)³ with x being in the interval (0; 20) because both numbers need to be positive.

In order to find the minimum, we need to find the zeros of the first derivative to see where the critical points are

f ' (x) = -3 x² +122 x-1200 = 0
-(-24 + x) (-50 + 3 x) = 0
x=24 or x= 50/3

24 is outside the given interval, so x=50/3
To check if that is really a minimum, we produce the second derivative

f '' (x) = -6x + 122

for x=50/3 the value of f '' is 22 which means that at x=50/3 we have a minimum

So the numbers are 50/3 and 20-50/3=60/3-50/3 = 10/3

x + y = 20
Minimize x^3 + y^2

y = 20 - x
F(x) = x^3 + (20 - x)^2

Find critical numbers of F(x):
F'(x) = 3x^2 + 2(20 - x)(-1)
= 3x^2 - 40 + 2x
= (x + 4)(3x - 10)

F'(x) = 0 or F'(x) DNE
No DNE critical numbers
x = {-4, 10/3}

x>0, therefore, x = 10/3 and y = 50/3.

Check using 1st or 2nd derivative test.