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Chapter 4 | Applications of Derivatives
Figure 4.64 A square with side length x inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.
Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V . Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is V = L · W · H , where L , W , and H are the length, width, and height, respectively. Step 4: From Figure 4.64 , we see that the height of the box is x inches, the length is 36−2 x inches, and the width is 24−2 x inches. Therefore, the volume of the box is V ( x ) = (36−2 x )(24−2 x ) x =4 x 3 −120 x 2 +864 x . Step 5: To determine the domain of consideration, let’s examine Figure 4.64 . Certainly, we need x >0. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for x over the open interval (0, 12). Since V is a continuous function over the closed interval [0, 12], we know V will have an absolute maximum over the closed interval. Therefore, we consider V over the closed interval [0, 12] and check whether the absolute maximum occurs at an interior point. Step 6: Since V ( x ) is a continuous function over the closed, bounded interval [0, 12], V must have an absolute maximum (and an absolute minimum). Since V ( x ) =0 at the endpoints and V ( x ) >0 for 0< x <12, the maximum must occur at a critical point. The derivative is V ′( x ) =12 x 2 −240 x +864. To find the critical points, we need to solve the equation 12 x 2 −240 x +864=0. Dividing both sides of this equation by 12, the problem simplifies to solving the equation x 2 −20 x +72=0. Using the quadratic formula, we find that the critical points are
2 −4(1)(72) 2
20±4 7
x = 20± (−20)
= 20± 112 2 =
2 =10±2 7.
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