# Calculus

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

• Calculus -

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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