odporko said:
1. Homework Statement
f(x,y) = x^(1/4)*y^(1/3), g(x,y) = 4*x +6*y=720
sorry i had a mistake before
2. Homework Equations
∇f=λ∇g
y^(1/3)/(4*x^(3/4))= 4*λ
x^(1/4)/(3*y^(2/3))= 6*λ
these are partial derivations, and we want to express y just with lambda and here is something that really terrifies me: 4096*((104976*y^(8/3)*λ)^(9/4))*λ, am i thinking right? or am i going wrong?
You can use the first equation to solve for y in terms of (x,λ), then substitute that y into the second equation to get an equation containing only x and λ. Solve for x in terms of λ, then substitute that x into the expression for y. You will end up with x and y expressed only in terms of λ, and the expressions are not too horrible:
x = \frac{2^{4/5}\, 3^{2/5}}{2304\, \lambda^{12/5}}, \;<br />
y = \frac{2^{4/5}\, 3^{2/5}}{2592\, \lambda^{12/5}}.
Substituting these into the equation for g you get a simple linear equation in the variable
\lambda^{12/5}, so you can solve it and pick the relevant real, positive root. This turns out to be
\lambda = \frac{5^{7/12}\, 3^{1/12}\, 7^{5/12}\, 2^{1/6}}{720} \doteq 0.009827792083. Now you can find x and y:
x = 540/7, \; y = 480/7, \; f(x,y) = \frac{540^{1/4}\, 7^{5/12} \,480^{1/3}}{7}<br />
\doteq 12.13030337. This is the constrained max.
The constrained min cannot be found by Lagrange multipliers. Just recognize that f(x,y) is not defined for x < 0 and y < 0, so we are restricted to the region {x ≥ 0, y ≥ 0}, and that f = 0 when x = 0 or y = 0. Thus, there are two constrained min points, at (x,y) = (0,720/6) and (x,y) = (720/4,0), both giving f = 0.
RGV
