Optimization prob: rect prism w/ square base, vol 414 in^3; minimize material needed

[cosmia]

Hello, I have an optimization problem that says "Someone is making a rectangular prism with a square base that has a volume of 414 cubic inches. Find the dimensions of the box that will minimize the amount of material needed (Surface Area) needed to create the box."
The optimization equation I used was SA = 2x^2 + 4xy and the constraint I used was v=x^2y. I solved for y in my constraint, plugged it back in the optimization equation, etc etc.

After all the work was done, the dimensions I ended up with were about 7.53 x 7.53 x 7.53 inches (these are rounded) for the length, width, and height. However, I'm not sure if I'm right. Can anyone help me?

 

[stapel]

Someone is making a rectangular prism with a square base that has a volume of 414 cubic inches. Find the dimensions of the box that will minimize the amount of material needed (Surface Area) needed to create the box.

The optimization equation I used was SA = 2x^2 + 4xy and the constraint I used was v=x^2y. I solved for y in my constraint, plugged it back in the optimization equation, etc etc.

I will guess that "SA" stands for "surface area", that "x" is the side-length for the square base (so the cross-sectional area of the prism is x^2, as is the area of each of the two ends), and that "y" is the length of the remaining dimension of the prism. Then the volume, "V", is given by:

. . . . .volume: V = x²y = 414

...and the total surface area, comprised of the two ends and the four other sides, is given by:

. . . . .surface area: SA = 2x² + 4xy

You then solved the "volume" equation for the simpler variable, being "y":

. . . . .414 = x²y

. . . . .414/x² = y

You then plugged this into the "surface area" equation:

. . . . .SA = 2x² + 4x(414/x²) = 2x² + 1,656/x = 2x² + 1,656x^-1

Then you took the derivative and found the critical point(s), etc.

(By the way, the above is how one shows one's work for others to examine.)

After all the work was done, the dimensions I ended up with were about 7.53 x 7.53 x 7.53 inches (these are rounded) for the length, width, and height. However, I'm not sure if I'm right. Can anyone help me?

We can't check your work because it hasn't been shown. You can graph the "surface area" equation to check for the minimizing value of x (and the minimum value of y = SA_min). In looking at your answer, I note that the volume you give (approximately 427 in³) is fairly far afield of the specified value, so I suspect that rounding errors are, at the least, involved somewhere. But, other than that, what "help" are you requesting?

Please be specific. Thank you!