Find the dimensions of the box so that the total cost of material is least.?

Question

A closed box with square base is to have a volume of 6912 cubic inches. The material for the top and bottom of the box is to cost 4 cents per square inch and the material for the sides is to cost 1 cent per square inch. Find the dimensions of the box so that the total cost of material is least. If x is the length of the side of the square base and y is the depth of the box, then give your answer as (x,y).

Answer 1

OK, first let's try to get that into a mathematical statement. The volume can be given as yx^2, since the box is square, and the cost is 8x^2 for the top and bottom (4 cents times two sides), and 4xy for the sides (1 cent times four sides). So the problem can be stated mathematically as yx^2 = 6912, c = 8x^2 + 4xy, minimize c.

The next step, I guess, would be to rearrange the equation to eliminate one variable and solve for c in terms of the other. So say y = 6912/x^2, then c = 8x^2 + (4*6912)x/x^2. That simplifies to c = 8x^2 + 27648x^-1. Now we have c in terms of x.

The next step is to find the value of x for which c is minimized. You don't say what math class you're in but the only way I know how to do this is with calculus, so I assume you are that advanced. First you have to take the derivative of the function. So the derivative of 8x^2 +27648x^-1 is 16x -27648x^-2.

Next, find all points for which c' = 0. Set 16x = 27648/x^2. X = 1728/x^2. x^3 = 1728. x = 3_/1728 = 12. So the only local min/max is at x = 12.

Now, you would ordinarily compare the value of the function at the local min/max with the value at the boundaries, but since this function is undefined at the boundaries,we can assume that x = 12 is the answer.

Going back to the first equation, yx^2 = 6912. That means 144y = 6912. y = 48.

The answer is (12, 48).

Answer 2

V= s * s * h (* for multiplication) V = h s^2 = 36,000 (i will tack on the measurements on the top.) So h = 36,000/s^2 P (for fee) = .03 s^2 + .08 sh i'm hoping this is in sq. inches - you were not particular related to the cost and, for this difficulty, that counsel is genuinely needed. P = .03 s^2 + .08s * (36,000/s^2) = .03 s^2 + 2,880/s dP ---- = .06 s -2,880 dS Set that equivalent to 0 and .06s = 2880 Multiply the two factors by way of a hundred 6s = 288000 s = 40 8,000 (That fee could desire to no longer be in step with sq. inch - is it?) If the cost have been in step with sq. foot the respond could be noticeably distinctive. Sorry, or in step with risk I made a mistake.

Answer 3

Okay x^2*y=6912.
And the cost is represented by (2*.04*x^2)+(4*.01*xy).
Now y=6912/(x^2).
Now substitute that in cost, and you have (2*.04*x^2)+(4*.01*x*(6912/(x^2))) = (2*.04*x^2)+((4*.01*6912)/x).

So, differineate it and you have (4*.01*x)+((-4*.01*6912)/(x^2)).
Now, set that to 0 to find x.