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Chapter 4 | Applications of Derivatives
We want to find the maximum possible area subject to the constraint that the total fencing is 100ft. From Figure 4.62 , the total amount of fencing used will be 2 x + y . Therefore, the constraint equation is 2 x + y =100. Solving this equation for y , we have y =100−2 x . Thus, we can write the area as A ( x ) = x · (100−2 x ) =100 x −2 x 2 . Before trying to maximize the area function A ( x ) =100 x −2 x 2 , we need to determine the domain under consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need x >0 and y >0. Since y =100−2 x , if y >0, then x <50. Therefore, we are trying to determine the maximum value of A ( x ) for x over the open interval (0, 50). We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function A ( x ) =100 x −2 x 2 over the closed interval ⎡ ⎣ 0, 50 ⎤ ⎦ . If the maximum value occurs at an interior point, then we have found the value x in the open interval (0, 50) that maximizes the area of the garden. Therefore, we consider the following problem: Maximize A ( x ) =100 x −2 x 2 over the interval ⎡ ⎣ 0, 50 ⎤ ⎦ . As mentioned earlier, since A is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, A ( x ) =0. Since the area is positive for all x in the open interval (0, 50), the maximum must occur at a critical point. Differentiating the function A ( x ), we obtain A ′( x ) =100−4 x . Therefore, the only critical point is x =25 ( Figure 4.63 ). We conclude that the maximum area must occur when x =25. Then we have y =100−2 x = 100 − 2(25) = 50. To maximize the area of the garden, let x =25 ft and y =50ft. The area of this garden is 1250ft 2 .
Figure 4.63 To maximize the area of the garden, we need to find the maximum value of the function A ( x ) =100 x −2 x 2 .
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