from xy^2 = 16
y^2 = 16/x
now sub that into
F = x^2 + y^2
= x^2 + 16/x
dF/dx = 2x - 16/x^2
= 0 for max/min
2x - 16/x^2 = 0
2x = 16/x^2
2x^3 = 16
x^3 = 8
x = cuberoot(8) = 2
if x = 2 , y^2 = 16/2 = 8 ---> y = ±√8 or ±2√2
then F = x^2 + y^2
= 4 + 8 = 12
by 2nd derivative test
F '' = 2 + 32/x^3 which is > 0 for x = 2
so when x = 2, F is a minimum of 12
Can someone please walk me through the steps, I am just not sure what to do next. thanks
Solve the optimization problem.
Minimize F = x^2 + y^2 subject to xy^2 = 16
I took the derivative F = 2x + 2Y but I don't know where to go from here.
1 answer
1 answer