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 ![195112 in^{3}](https://tex.z-dn.net/?f=195112%20in%5E%7B3%7D)
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}](https://tex.z-dn.net/?f=V%3D%28w%5E%7B2%7D%29%28174-2w%29%5C%5CV%3D174w%5E%7B2%7D-2w%5E%7B3%7D)
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)](https://tex.z-dn.net/?f=V%27%3D348w-6w%5E%7B2%7D%20%5C%5CV%27%3D%28348-6w%29%28w%29)
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](https://tex.z-dn.net/?f=0%3D%28348-6w%29%28w%29%5C%5Cw_1%3D0%5C%5Cw_2%3D348%2F6%3D58)
The first value (
) will leave us with a 'height-only box', so the answer must be ![w=58 in](https://tex.z-dn.net/?f=w%3D58%20in)
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}](https://tex.z-dn.net/?f=195112%20in%5E%7B2%7D)