Solving Lagrange Multiplication Question

Question

Optimize $f(x,y,z)=25x^\frac 25y^\frac 15z^\frac15$ with the constraint $24000=80x+12y+10z$

New to Lagrange, I have no idea how to solve this

Answer 1

The gradient of the function is given by: $\bigtriangledown$ f= $\langle\frac{10y^\frac15z^\frac15}{x^\frac35},\frac{5x^\frac25z^\frac15}{y^\frac45},\frac{5x^\frac25y^\frac15}{z^\frac45}\rangle$
The gradient of the constraint is given by $\bigtriangledown$ g= $\langle80,12,10\rangle$

Since $\bigtriangledown f=\lambda \bigtriangledown g$
$80\lambda=\frac{10y^\frac15z^\frac15}{x^\frac35}$ which we can simplify as $\lambda=\frac{10y^\frac15z^\frac15}{80x^\frac35}$

$12\lambda=\frac{5x^\frac25z^\frac15}{y^\frac45}$ which we can simplify as $\lambda=\frac{5x^\frac25z^\frac15}{12y^\frac45}$

$10\lambda=\frac{5x^\frac25y^\frac15}{z^\frac45}$ which we can simplify as $\lambda=\frac{5x^\frac25y^\frac15}{10z^\frac45}$

Since we have all of our terms equivalent to $\lambda$, we can set the terms equivalent to each other and solve for a variable. We can begin by setting the first two terms equivlent to each other.
$\frac{10y^\frac15z^\frac15}{80x^\frac35}=\frac{5x^\frac25z^\frac15}{12y^\frac45}$
Some cross multiplication and some algebra leads us to the fact that:
$120yz^\frac15=400xz^\frac15$
$x=\frac{120}{400}y$
Since we now have x in terms of y, we need z in terms of y so that we can plug it back into our original constraint. So lets solve for z in terms of y by setting the last two terms equivalent to each other, like so:
$\frac{5x^\frac25z^\frac15}{12y^\frac45}=\frac{5x^\frac25y^\frac15}{10z^\frac45}$
Some cross multiplication and some algebra leads us to the fact that:
$50x^\frac25z=60x^\frac25y$
$z=\frac{60}{50}y$
We now have x and z in terms of y, and since y is already in terms of y, we can plug everything into our original constraint.
$80(\frac{120}{400}y)+12y+10(\frac{60}{50})y=24000$
$24y+12y+12y=24000$
$y=500$
Now that we have a value for y, and we have x and z in terms of y, we can just plug in y and find the values of x and z.
$x=\frac{120}{400}(500)=150$
$y=500$
$z=\frac{60}{50}(500)=600$
Plugging the x,y,z we found into our original function
$f(150,500,600)=25(150)^\frac25(500)^\frac15(600)^\frac15=2311.00435152$
Looking at our function, there is no lower bound therefore our maximum is given by $f(150,500,600)=2311.00435152$