Calculus - Min max problem - having trouble working out the derivatives:?

Question

Find 2 numbers whose product is 36 and the sum of their cubes is a minimum.
Any help appreciated.

Answer 1

hi

x*y = P

f(x;y)=x^3 + y^3
G(x) = x^3 + P^3 / x^3

G' = 3 x^2 - 3 P^3/x^4 = 0

x^6 = P^3

x^2 = P

x=y=VP

then

x=y=6

bye

Answer 2

x*y=36
z= x^3+y^3 minimum
z= x^3 +36^3/x^3
z´= 3x^2 --3*36^3/x^4 = 0
x^6-6^6 =0 so x=+-6 as 36=6^2
The sign of the derivative is ++++++ -6--------6++++++
so at x= 6 we have a minimum
y=6
and z=2*6^3