# What is the largest volume box that can be shipped?

To send a box, a courier service requires that the length plus twice the width plus the double the height does not exceed 280 centimeters. What is the largest volume box that can be shipped

I know that volume of a box is $$(x)(y)(z)$$ and I know that $$2x+2y<=280cm$$ and I think its using Lagrange multiplier but I don't know how to exactly use that is <=280 I used to use just "="

• The inequality is $x+2y+2z\le 280$. And yes, use $=$ Oct 29, 2020 at 17:59
• Then I should Use x+2y+2z=280 as restriction? Oct 29, 2020 at 18:01
• Yes you should. Oct 29, 2020 at 18:02
• @KevinDuran If $x+2y+2z < 280$ then you can always add a tiny bit more to any one, or all of $x,y,z$ to make it equal to $280$, also making a bigger volume box Oct 29, 2020 at 18:03

$$V = x y z = (280 - 2 y - 2 z) y z$$. But clearly (by symmetry) the solution occurs when $$y = z$$. To see this: there is nothing in the problem that distinguishes the depth $$z$$ and the width $$y$$. We could relabel $$y$$ with $$z$$ and the problem would be identical... as is clear from the functional form: $$(280 - 2 y - 2 z) y z$$. Given that there is a single optimum, that demands that $$y = z$$ at that optimum.
Thus $$V = (280 - 4y) y^2$$. Set $$\frac{dV}{dy}= 2 (280 - 4 y) y - 4 y^2 = 0$$ to find $$y = z = 140/3$$, and resubstitute to find $$x = 280/3$$. Then $$V = 5488000/27$$.