You are building a barn, with no floor, in the shape of a rectangular box with a square base. The roof material costs $\$19/\text{m}^2$, the sides and back material costs $\$13/\text{m}^2$, and the front material costs $\$20/\text{m}^2$. The volume of the barn will be $16,000 \text{ m}^ 3$. What dimensions minimize the total cost? (answer dimensions from the smallest to the largest and round your answer to the second decimal point)
I set up my equation like this $19x^2+59xy$ with a constraint at $x^2y=16000$
I got $y=38/59$ and $x=\sqrt{9444000/38},-\sqrt{9444000/38}$
I am not too sure on what I did wrong.
If $x$ is the side of the floor, and $y$ the height, and the roof is flat, then indeed we are minimizing $f(x,y)=19x^2+59xy$ subject to the constraint $x^2 y=16000$.
This is easily transformed into a $1$ variable problem by substituting $y=\frac{16000}{x^2}$ for $y$ in $19x^2+59xy$. But I gather you want to use Lagrange multipliers.
Taking partial derivatives, we get the two equations $$38x+59y=2\lambda xy\quad\text{and} \quad 59x=\lambda x^2.$$ Since $x\ne 0$, we have $\lambda=\frac{59}{x}$. Substituting in the first equation, we get $38x=59y$.
That gives $x=\sqrt[3]{(59)(16000)/38}$. The value of $y$ is $\frac{38}{59}x$.
Remark: It looks as if you had everything right except for a slip at the end.