0
$\begingroup$

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 "="

$\endgroup$
4
  • $\begingroup$ The inequality is $x+2y+2z\le 280$. And yes, use $=$ $\endgroup$
    – David P
    Oct 29, 2020 at 17:59
  • 1
    $\begingroup$ Then I should Use x+2y+2z=280 as restriction? $\endgroup$ Oct 29, 2020 at 18:01
  • $\begingroup$ Yes you should. $\endgroup$ Oct 29, 2020 at 18:02
  • $\begingroup$ @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 $\endgroup$
    – David P
    Oct 29, 2020 at 18:03

1 Answer 1

0
$\begingroup$

$V = x y z = (280 - 2 y - 2 z) y z$. 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 $V = 5488000/27$.

$\endgroup$
1
  • 2
    $\begingroup$ This "clearly" is not so clear. $\endgroup$ Oct 29, 2020 at 19:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .