# Lagrange Multiplier

• Mar 29th 2009, 07:25 AM
Macleef
Lagrange Multiplier
See attach for the question. My answer is incorrect. Where did I go wrong?

$\displaystyle V = x^2y = 900$

$\displaystyle SA = 6xy$

$\displaystyle f_x = 6y$

$\displaystyle f_y = 6x$

$\displaystyle g_x = 2xy$

$\displaystyle g_y = x^2$

$\displaystyle 6x = 2xy \lambda$

$\displaystyle 6x = x^2 \lambda$

$\displaystyle \frac{3x}{xy} = \frac{6x}{x^2}$

$\displaystyle 3x^3 = 6x^2y$

$\displaystyle \frac{1}{2}x = y$

$\displaystyle x^2 (\frac{1}{2}x) = 900$

$\displaystyle x = 12.1644$

$\displaystyle y = 6.0822$
• Mar 29th 2009, 08:03 AM
Jester
Quote:

Originally Posted by Macleef
See attach for the question. My answer is incorrect. Where did I go wrong?

$\displaystyle V = x^2y = 900$

$\displaystyle SA = 6xy$

$\displaystyle f_x = 6y$

$\displaystyle f_y = 6x$

$\displaystyle g_x = 2xy$

$\displaystyle g_y = x^2$

$\displaystyle 6x = 2xy \lambda$ (**)

$\displaystyle 6x = x^2 \lambda$

$\displaystyle \frac{3x}{xy} = \frac{6x}{x^2}$

$\displaystyle 3x^3 = 6x^2y$

$\displaystyle \frac{1}{2}x = y$

$\displaystyle x^2 (\frac{1}{2}x) = 900$

$\displaystyle x = 12.1644$

$\displaystyle y = 6.0822$

I believe here (**)