I REALLY REALLY NEED HELP. I WILL MARK BRAINLIEST FOR THE BEST ANSWER!!! Question: What are the dimensions of a box with minimal surface area that has a volume of 64000 cm^3? AND REMEMBER, PLEASE SHOW ALL WORK NECESSARY. THANKS. God Bless :)

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  40 cm × 40 cm × 40 cm

Step-by-step explanation:

Since this is in the high-school math forum, we assume that you're not intended to use calculus to determine the answer. All three dimensions are open for variation, so we'll try to use some logic to determine a reasonable solution.

The dimensions of a box are interchangeable with respect to their effect on surface area. You can see this in the area formula:

  A = 2(LW +LH +WH)

That is, no particular dimension makes more or less contribution to surface area than any other. This suggests that the minimum area will be obtained when each dimension is equal to the others. That is, a cube has the minimum surface area for a given volume.

The volume of a cube in terms of its edge dimension s is ...

  V = s³

So, the edge dimension is ...

  s = ∛V = ∛(64000 cm³) = 40 cm

The dimensions of the box are 40 cm × 40 cm × 40 cm.

Calculus method of Lagrange Multipliers

We want to minimize A=2(LW +WH +LH) subject to the constraint that LWH = 64000. We can write the Lagrangian as ...

  L = 2(LW +WH +LH) +λ(LWH -64000)

We want to set all of the partial derivatives of L to zero.

  dL/dL = 0 = 2(W+H) +λWH

  dL/dW = 0 = 2(L+H) +λLH

  dL/dH = 0 = 2(L+W) +λLW

  dL/dλ = 0 = LWH -64000

Solving the first two equations for λ and setting the results equal, we have ...

  -2(W+H)/(WH) = λ = -2(L+H)/(LH)

Multiplying by H/(-2) gives ...

  1 +H/W = 1 +H/L   ⇒   W = L

Similarly, solving the 2nd and 3rd equations for λ and setting the results equal, we have ...

  -2(L+H)/(LH) = λ = -2(L+W)/(LW)

  1 +L/H = 1 +L/W   ⇒   H = W

So, now we know that L = W = H, which is the assumption we started with in our "logical" answer.

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