# Thread: Lagrange Multiplier Maximization Problem

1. ## Lagrange Multiplier Maximization Problem

This is an example problem our professor gave us but after getting the partials, it jumps right to the answer for x and y. Is solving this problem just a matter of doing some algebra to solve for x and y using the three resulting equations from the partial derivatives?

Maximize 2x-5y subject to x^4 + y^4 = 1

F(x,y,L) = 2x-5y+L(x^4+y^4-1)

partial of F with respect to x, 2+4*L*x^3=0
partial of F with respect to y, -5+4*L*y^3=0
partial of F with respect to L, x^4 + y^4 -1=0

2. Originally Posted by soma
This is an example problem our professor gave us but after getting the partials, it jumps right to the answer for x and y. Is solving this problem just a matter of doing some algebra to solve for x and y using the three resulting equations from the partial derivatives?

Maximize 2x-5y subject to x^4 + y^4 = 1

F(x,y,L) = 2x-5y+L(x^4+y^4-1)

partial of F with respect to x, 2+4*L*x^3=0
partial of F with respect to y, -5+4*L*y^3=0
partial of F with respect to L, x^4 + y^4 -1=0
The first equation is the same as 4Lx^3= -2, and the second 4Ly^3= -5. Since you don't need to know the value of L, divide one equation by the other: $\frac{x^3}{y^3}= \frac{5}{3}$ so that $x^3= \frac{5}{3}y^3$ and $x= \left(\frac{5}{3}\right)^{1/3}y$. Putting that into the constraint equation, $\left(\frac{5}{3}\right)^{4/3}y^4+ y^4= 1$.