Calculus
posted by help .
Q. Find the minimum value of Q=x^2y subject to the constraint 2x^2+4xy=294.
Its the derivative method.

from your constraint equation
4xy = 2942x^2
y = (2942x^2)/(4x)
so Q = x^2 (2942x^2)/(4x)
= 147x/2  x^3 /2
dQ/dx = 147/2  (3/2)x^2 = 0 for a min of Q
3x^2 /2 = 147/2
3x^2 = 147
x^2 = 49
x= ± 7
then y = ±7
if x=7, then y=7
if x = 7, then y = 7