Math Problem Statement

The volume of an open-top rectangular box is 4500 cc (cubic centimeters). The length of the rectangular base of the box is twice the width. What height will make the surface area as small as possible?

Solution

To solve this problem, we need to minimize the surface area of an open-top rectangular box while keeping its volume constant at 4500 cubic centimeters (cc).

Step 1: Define the dimensions

Let:

  • ww be the width of the box.
  • ll be the length of the box, and we are told l=2wl = 2w (the length is twice the width).
  • hh be the height of the box.

Step 2: Volume equation

The volume VV of a rectangular box is given by: V=l×w×hV = l \times w \times h Substitute l=2wl = 2w: V=2w×w×h=2w2hV = 2w \times w \times h = 2w^2h We are told the volume is 4500 cc, so: 2w2h=45002w^2h = 4500 From this, solve for hh: h=45002w2=2250w2h = \frac{4500}{2w^2} = \frac{2250}{w^2}

Step 3: Surface area equation

The surface area SS of an open-top box (no top) is given by: S=l×w+2(l×h)+2(w×h)S = l \times w + 2(l \times h) + 2(w \times h) Substitute l=2wl = 2w: S=2w×w+2(2w×h)+2(w×h)S = 2w \times w + 2(2w \times h) + 2(w \times h) Simplify: S=2w2+4wh+2wh=2w2+6whS = 2w^2 + 4wh + 2wh = 2w^2 + 6wh Now substitute h=2250w2h = \frac{2250}{w^2} from Step 2: S=2w2+6w×2250w2=2w2+13500wS = 2w^2 + 6w \times \frac{2250}{w^2} = 2w^2 + \frac{13500}{w}

Step 4: Minimize the surface area

To minimize the surface area, take the derivative of SS with respect to ww and set it equal to zero: dSdw=4w13500w2\frac{dS}{dw} = 4w - \frac{13500}{w^2} Set dSdw=0\frac{dS}{dw} = 0 and solve for ww: 4w=13500w24w = \frac{13500}{w^2} Multiply both sides by w2w^2: 4w3=135004w^3 = 13500 Solve for ww: w3=135004=3375w^3 = \frac{13500}{4} = 3375 w=33753=15cmw = \sqrt[3]{3375} = 15 \, \text{cm}

Step 5: Solve for height

Now that we have w=15cmw = 15 \, \text{cm}, substitute this into the expression for hh: h=2250w2=2250152=2250225=10cmh = \frac{2250}{w^2} = \frac{2250}{15^2} = \frac{2250}{225} = 10 \, \text{cm}

Conclusion:

The height that will make the surface area as small as possible is h=10cmh = 10 \, \text{cm}.


Would you like further details on any step or more explanation?

Here are 5 related questions to explore further:

  1. How would the problem change if the box had a top?
  2. Can you calculate the minimum surface area for the box?
  3. What happens if the length is three times the width instead of twice?
  4. How does the surface area function change if the box’s volume is different?
  5. How would this problem change if you wanted to minimize cost instead of surface area?

Tip: For optimization problems, always double-check that your critical points minimize (or maximize) the function by using the second derivative or other methods.

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Math Problem Analysis

Mathematical Concepts

Optimization
Algebra
Geometry

Formulas

Volume of a rectangular box: V = l * w * h
Surface area of an open-top box: S = lw + 2lh + 2wh
Derivative of surface area for minimization: dS/dw

Theorems

Optimization via first derivative

Suitable Grade Level

Grades 10-12