Chapter 4 | Applications of Derivatives
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Example 4.37 Minimizing Surface Area
A rectangular box with a square base, an open top, and a volume of 216 in. 3 is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area? Solution Step 1: Draw a rectangular box and introduce the variable x to represent the length of each side of the square base; let y represent the height of the box. Let S denote the surface area of the open-top box.
Figure 4.69 We want to minimize the surface area of a square-based box with a given volume.
Step 2: We need to minimize the surface area. Therefore, we need to minimize S . Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is x · y . The area of the base is x 2 . Therefore, the surface area of the box is S =4 xy + x 2 . Step 4: Since the volume of this box is x 2 y and the volume is given as 216 in. 3 , the constraint equation is x 2 y =216. Solving the constraint equation for y , we have y = 216 x 2 . Therefore, we can write the surface area as a function of x only:
⎛ ⎝ 216 x 2
⎞ ⎠ + x 2 .
S ( x ) =4 x
Therefore, S ( x ) = 864 x + x 2 .
Step 5: Since we are requiring that x 2 y =216, we cannot have x =0. Therefore, we need x >0. On the other hand, x is allowed to have any positive value. Note that as x becomes large, the height of the box y becomes correspondingly small so that x 2 y =216. Similarly, as x becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval (0, ∞). Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval (0, ∞). Step 6: Note that as x →0 + , S ( x )→∞. Also, as x →∞, S ( x )→∞. Since S is a continuous function
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