The square-based box with the greatest volume under the condition that the sum of length, width, and heigth does not exceed 174 in is a cube with each edge of 58 in and a volume of $195112 in^{3}$

Step-by-step explanation:

For this problem we have two constraints, that are as follows:

1) Sum of length, width, and heigth not exceeding 174 in

2) Lenght and width have the same measure (square-based box)

We know that volume is equal to the product of all three edges, and with the two conditions into account we have the next function:

$V=(w^{2})(174-2w)\\V=174w^{2}-2w^{3}$

The interval of interest of the objective function is [0, 87]

This problem requieres that we maximize the function that defines the volume. We start calculating the derivative of the function, wich is:

$V'=348w-6w^{2} \\V'=(348-6w)(w)$

We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.

So we find the value in wich the derivative equals zero:

$0=(348-6w)(w)\\w_1=0\\w_2=348/6=58$

The first value ( $w=0$ ) will leave us with a 'height-only box', so the answer must be $w=58 in$

The value is between the interval of interest.

And, once we solve for the constraints, we have that:

Lenght = Width = Heigth = 58 in

Volume = $195112 in^{2}$

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3 years ago