Answer to Question #250515 in Calculus for Susan

Question

The combined dimensions of a passenger's carry on bag may not exceed 45 inches. That is the length, width and height of the suitcase can add up to at most 45 inches. Find the maximum volume of a rectangular suitcase that meets these requirements.

Answer 1

The point is to maximize the value V = xyz, where x, y, z - lenght, width, and height, provided that x + y + z = 45

Firstly we will find x and y provided that x + y = a, at which S is maximized, where:

S = xy, where S - square of the two-dimensional base of the rectangle

x + y = a (a < 45)

y = a - x

$S = x(a-x) = ax - x^{2}$, $x\in (0;a)$

$S' = a-2x = 0$

$x = {\frac a 2}$

$y = {\frac a 2}$

$S={\frac {a^{2}} 4}$

where $x = {\frac a 4}$ or $x = {\frac {3a} 4}$ $S = {\frac{3a^{2}} {16}} < {\frac {a^{2}} 4}$, so $x = {\frac a 2}$ is the point where S is maximized

So, we found out that x = y. Now the task is to maximize $V = x^{2}z$ provided that 2x + z = 45

z = 45 - 2x

$V = x^{2}(45-2x) = 45x^{2} - 2x^{3}$ , $x \in (0;{\frac {45} 2})$

$V'= 90x - 6x^{2} = 0$

x = 0 or x = 15

x = 0 doesn't satisfy the conditions

x = 15 satisfies the conditions

z = 45 - 30 = 15

x = 15, y = 15, z = 15

The suitcase has the form of the cube

The maximume volume is $V = 15^{3} = 3375$ square inches.