Chapter 4 | Applications of Derivatives
447
Maximizing the Area of an Inscribed Rectangle
A rectangle is to be inscribed in the ellipse
x 2
y 2 =1.
4 +
What should the dimensions of the rectangle be to maximize its area? What is the maximum area?
Solution Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let L be the length of the rectangle and W be its width. Let A be the area of the rectangle.
Figure 4.68 We want to maximize the area of a rectangle inscribed in an ellipse.
Step 2: The problem is to maximize A . Step 3: The area of the rectangle is A = LW . Step 4: Let ( x , y ) be the corner of the rectangle that lies in the first quadrant, as shown in Figure 4.68 .We can write length L =2 x andwidth W =2 y . Since x 2 4 + y 2 =1 and y >0, we have y = 1− x 2 4 . Therefore, the area is A = LW = (2 x ) ⎛ ⎝ 2 y ⎞ ⎠ =4 x 1− x 2 4 =2 x 4− x 2 . Step 5: From Figure 4.68 , we see that to inscribe a rectangle in the ellipse, the x -coordinate of the corner in the first quadrant must satisfy 0< x <2. Therefore, the problem reduces to looking for the maximum value of A ( x ) over the open interval (0, 2). Since A ( x ) will have an absolute maximum (and absolute minimum) over the closed interval [0, 2], we consider A ( x ) =2 x 4− x 2 over the interval [0, 2]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval. Step 6: As mentioned earlier, A ( x ) is a continuous function over the closed, bounded interval [0, 2]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints x =0 and x =2, A ( x ) =0. For 0< x <2, A ( x ) >0. Therefore, the maximum must occur at a critical point. Taking the derivative of A ( x ), we obtain
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