Best dimensions for maximum surface area

[bjgawp]

I was doing some math problems involving surface area and maximum dimensions and then I wondered:
Suppose you are given the surface area of a rectangular box but none of its dimensions. Is it possible to find the best dimensions (x,y,z) that would give the maximum volume of the box? I was thinking of something along the lines of making a graph and finding its maximum value ... Ex. 400cm² = 2xy + 2xz + 2yz. Possible? Unless there are other ways of doing it...

 

[HallsofIvy]

Make a graph of what? Not the surface area function that you give: you want to maximize the volume: V= xyz.

The standard way of "maximizing" (or "minimizing") a function is to find its derivative and set that equal to 0. Since here you have a function of three variables (x,y,z) and an additional constraint, there are two ways to do it.

1) Use the constraint to reduce the number of variables: 2xy+ 2xz+ 2yz= 400 so 2(x+y)z= 400- 2xy so z= (200-xy)/(x+y) and then V= xy(200-xy)/(x+y). Differentiate that with respect to x and y (partial derivatives) and set them equal to 0. You can probably see those are going to be messy derivatives!

2) The Lagrange Multiplier method: At a maximum (or minimum) value, the gradient of the function must be a multiple of the gradient of the constraint. Here the function to be maximized is V= xyz. grad V= yzi+ xzj+ xyk. The constraint function is U= 2xy+ 2xz+ 2yz= 400 and grad U= (2y+ 2z)i+ (2x+2z)j+ (2x+2y)k. One is a multiple of the other if [itex]yz= \lambda(2y+ 2z)[/itex], [itex]xz= \lamba(2x+2z)[/itex], and [itex]xy= \lambda(2x+ 2y)[/itex]. That [itex]\lambda[/itex] is the "Lagrange Multiplier".
If you eliminate [itex]\lambda[/itex] from those, you can get immediately that x= y= z: a cube will maximize volume for given surface area.

 

[tacman]

Yes, it is possible to find the best dimensions for maximum surface area using mathematical techniques such as optimization. In this case, you would need to use the given surface area formula and set it equal to the maximum volume formula for a rectangular box, which is V = lwh. Then, using calculus or other optimization methods, you can find the values for x, y, and z that would give the maximum volume. However, it is important to note that there may be multiple solutions for the dimensions that would result in the same maximum volume. Additionally, you would need to consider any constraints or limitations, such as the dimensions needing to be positive numbers or within a certain range. Overall, it is definitely possible to find the best dimensions for maximum surface area using mathematical techniques.