Calculus Optimization problem?

Question

A block of ice maintains the shape of a cube as it melts at the constant rate of 3 cubic inches per minute. Find the rate at which the surface area of the cube is changing when the side of the cube is 12 inches long. Hint: Recall that a cube has six faces.

Please show all work.

Sorry it is a related rates problem. My bad.

Answer 1

Here's what we know so far:
1. Let x = the length of one of the sides of the cube in inches.
2. The volume of the cube is x^3.
3. The rate of change of volume is d(x^3)/dt = 3.
4. The surface area of one of the sides is x^2.
5. The surface area of the entire cube (six sides) is 6*x^2.
6. The rate of change of surface area is d(6*x^2)/dt. (This is what you're trying to solve for.)
Here's the solution:
7. Use the chain rule on (3): d(x^3)/dt = d(x^3)/dx*dx/dt = 3*x^2*dx/dt = 3; or x^2*dx/dt = 1.
8. Evaluate (7) with x = 12: 12^2*dx/dt = 1 or dx/dt = 1/144.
9. Use the chain rule on 6: d(6*x^2)/dt = d(6*x^2)/dx * dx/dt = 12*x * dx/dt.
10. Substitue (8) into (9): d(6*x^2)/dt = 12*x * 1/144 = x/12.
11. Evaluate (10) with x = 12: d(6*x^2)/dt = 12/12 = 1. Units are inches^2/minute. Answer is 1 sq. inch per minute.

Answer 2

a million) it really is basic; ==> First settle on the purpose of the placement; right here it really is paper dimensions, so as that its section will be minimum; then next comes lower than what constraints this must be finished? ==> properly, right here on the selected paper, you're leaving some margins throughout and arriving on the area for printing, it really is given some fixed cost. 2) Now, we are sparkling what we pick; the thanks to proceed for progression mathematical equations, so as that it will be solved for procuring the right answer: 3) right here we are given the printing section as 50 squarewherein is consistent; for this reason enable us initiate from this information; ==> enable the measurement of the printing section be x in (perfect smart) by y in (width smart) ==> Printing section = xy = 50; == y = 50/x -------(a million) 4) there's a margin of four in each and every at perfect and bottom are provided; for this reason average perfect of the paper is "x + 8" in; further a margin of two in is equipped on both aspects; ==> average width = y + 4 in 5) for this reason the area of the paper is = (x+8)(y+4) 6) Substituting for y from equation (a million), A (x+8)(50/x + 4) = 50 + four hundred/x + 4x + 32 7) for this reason the function to be minimized is: A = 80 2 + 4x + four hundred/x 8) Differentiating this, A' = 0 + 4 - 2 hundred/x^2 = 4 - four hundred/x^2 9) Equating A' = 0, x^2 = 100; ==> x = +/- 10 in 10) yet a measurement won't be able to be negative, for this reason we evaluate purely + 10; So x = 10 in and y = 5 in 11) although we favor to substantiate,no matter if that's minimum or optimal; for which we can save on with 2d by-product attempt; So again differentiating, A'' = 800/x^3; at x = 10, A'' is > 0; for this reason it really is minimum therefore we finish for the printing the area to be 50 squarein, lower than the given constraints of margins, the outer length of the paper should be 18 in by 9 in; So outer section is = 162 squarein. wish you're defined; have a impressive time.

Answer 3

The only thing worth pointing out is that the surface area of the cube is 6x^2, and not 12x^2 as the first poster indicated.

I'm sure that he can correct his error and produce a correct solution, because everything else that he did is entirely correct.

Cheers!

Answer 4

Let side of cube be x

A = 12x^2 or dA/dt = 24x dx/dt

V = x^3
So dV/dt = 3x^2 dx/dt

x = 12, dV/dt = 3 So, dx/dt = 1/144 in/min.

Thus dA/dt = 24*12 * 1/144 sq in/min = 2 sq in/min.

Answer 5

I don't do homework, but to help, this a related rates problem. The volume is changing with respect to time. That's a derivative or a rate.
and
V=s^3
That should get you going.