Answer:
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 [tex]195112 in^{3}[/tex]
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:
[tex]V=(w^{2})(174-2w)\\V=174w^{2}-2w^{3}[/tex]
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:
[tex]V'=348w-6w^{2} \\V'=(348-6w)(w)[/tex]
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:
[tex]0=(348-6w)(w)\\w_1=0\\w_2=348/6=58[/tex]
The first value ([tex]w=0[/tex]) will leave us with a 'height-only box', so the answer must be [tex]w=58 in[/tex]
The value is between the interval of interest.
And, once we solve for the constraints, we have that:
Lenght = Width = Heigth = 58 in
Volume = [tex]195112 in^{2}[/tex]
