Question

Fly-by-Night Airlines has a peculiar rule about luggage: The length l and width w of a bag must add up to at most 57 inches, and the width w and height h must also add up to at most 57 inches. What are the dimensions of the bag with the largest volume that Fly-by-Night will accept?

Answer 1

Maximize LxHxW

Constraint L+W=57 and W+H=57

in this case L must equal H since the sums are both 57

So the maximization becomes L²xW

the constraint is L+W=57

Solving the constraint for W yields

W=57-L

place this into the maximization eqn

$f(L)=L^{2}(57-L)= -L^{3}+57L^{2}$

now we take the derivative of this equation and determine the critical points (where the derivative=0)

derivative

$f'(L)=-3L^{2}+114L=0$

use the quadratic equation to solve

this yields L=0 and L=38

obviously L cannot = 0

solve for the other variables

if L=38 then H=38

L+W=57

W=57-L=57-38=19

L=38

H=38

W=19

Vmax=38*38*19=27436 cubic inches

Answer 2

The lenth $l$ and width $w$ of a bag must add up to at most $57$ inches.

$\Rightarrow l+w\le 57$  $\Rightarrow l\le 57-w...(1)$

The height $h$ and width $w$ of a bag must add up to at most $57$ inches.

$\Rightarrow h+w\le 57$  $\Rightarrow h\le 57-w...(2)$

Now, we have to maximize the volume, $lwh$, of such a bag.

$lwh\le (57-w)w(57-w)$ [ Using $(1), (2)$]

$\Rightarrow lwh\le w(57-w)^2=w(w^2-114w+3249)=w^3-114w^2+3249w$

Now, we have to maximize $g(w)=w^3-114w^2+3249w$ to get the answer.

$g'(w)=3w^2-2\times114w+3249=3(w^2-76w+1083)=3(w-57)(w-19)$

Now, when, $g(w)$ attains extremum, then, $g'(w)=0\Rightarrow 3(x-19)(x-57)=0\Rightarrow w=19,57$

$g''(w)=3(2w-76)$

$g$ attains maximum when $g''(w)<0$.

We check that, $g''(19)=3(2\times 19-76)=-3\times 38=-114<0$

and, $g''(57)=3(2\times 57-76)=3\times 38=114>0$

$\Rightarrow g$ attains maximum when $w=19$

$\Rightarrow lwh\le 19^3-114\times 19^2+3249\times 19=27436$

The dimensions of the bag with largest volume that fly-by-night will accept are

$l=57-19=38$ inches, $w=19$ inches, $h=57-19=38$ inches.