A box with a rectangular base, whose length is twice its width, is to have a closed top. The area of the material in the box is to be $192in^2$. What should the dimensions of the box be in order to have the largest possible volume?
$$l = 2w$$
So,
$$S= 2(lw)+ 2(wh)+ 2(lh)
\\ = 2(2w^2)+ 2(wh)+2(2wh)
\\ = 4w^2+2wh+4wh
\\ = 4w^2 +6wh$$
$$V=lwh
\\ = 2w \cdot w \cdot h
\\ = 2w^2 \cdot h$$
Now,
$${dV \over dw} = 4w\cdot {dh \over dw}
\\ 0= 4w \cdot {dh \over dw}
\\ {dh \over sw} = -4w$$
and
$${dS \over dw} = 4w^2 +6wh
\\ = 8w + 6 \cdot {dh \over dw}
\\ = 8w + 6(-4w)
\\ = -16w$$
I think I'm lost. I don't know what to do next. Could someone help find the solution? Thanks!
Best Answer
Hint
If you wish to apply Lagrange Multiplier method, after simplification of V,S
$$S1 = S/2 , V1 = V/2 $$
$$ \frac{\partial S1_h} {\partial V1_h} =\frac{\partial S1_w} {\partial V1_w}$$
$$ \rightarrow w/h = 3/2. $$