Posted by **help** on Thursday, March 15, 2012 at 7:22pm.

Q. Find the minimum value of Q=x^2y subject to the constraint 2x^2+4xy=294.

Its the derivative method.

- Calculus -
**Reiny**, Thursday, March 15, 2012 at 8:22pm
from your constraint equation

4xy = 294-2x^2

y = (294-2x^2)/(4x)

so Q = x^2 (294-2x^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

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