Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.7
Find the dimensions of the largest rectangle that can be inscribed in the upper half of the ellipse 5 x2+12 y2=60.
Solution
Analysis
Figure 3.8.7(a) is an animation that displays rectangles inscribed in the upper half of the given ellipse, and shows the area of each such rectangle. At the top of the animation, the value of x, half the width of the rectangle, is displayed.
If the figure is visible, simply click on it to access the animation toolbar, in the center of which there is a slider that controls the animation.
If the figure is not visible, click on the button "Figure 3.8.13" to activate the animation code.
The 101 frames in the animation subdivide a=12, the semi-major axis of the ellipse, into 100 equal divisions, so that each frame increments x by 0.034641.
Within the resolution supported by this animation, the largest rectangle has width twice x=2.4595, and area 7.7460. (See frame 72 in the animation.)
Animation: Rectangle in ellipse
The equation of the ellipse defines y=yx implicitly. Since the rectangle has its two upper corners on the ellipse, the height of the rectangle is yx, and the width of the rectangle is 2 x. Hence, the objective function for this example is the area of the ellipse, namely,
Ax=2 x yx=2 x 51−x2/12
Numeric Solution
Numeric optimization as a two-variable problem
Write a sequence consisting of the objective function and the constraint equation. Press the Enter key.
Context Panel: Optimization≻Maximize (local)
2 x y,5 x2+12 y2=60
2⁢x⁢y,5⁢x2+12⁢y2=60
→maximize
7.74596669241495750,x=2.44948974278320,y=1.58113883008420
The return consists of a list with two objects. The first object is the optimal value of the objective function (area); the second, a list of the parameter values giving this extreme value.
Numeric optimization as a single-variable problem
The graph of Ax in Figure 3.8.7(b) shows an absolute maximum of about 7 for x≐2.5. The graph itself can be probed for a numeric estimate of the maximum point. Alternatively:
Write 2 x 51−x2/12 Context Panel: Optimization≻ Optimization Assistant
See Figures 3.8.7(c) and 3.8.7(d).
Click to launch the Optimization Assistant with all the data in place.
Figure 3.8.7(b) Graph of Ax
Figure 3.8.7(c) Optimization Assistant solution
Figure 3.8.7(d) Constraints and Bounds window
Naive attempts to obtain the maximum of Ax with the Context Panel or even with the Optimization Assistant, fail. The implied constraint x∈0,12=0,23 must be included in the calculation. The only syntax-free way to add this to the calculation is through the Optimization Assistant.
In the Optimization Assistant, click the Edit button to the right of "Constraints and Bounds." The Constraints window (Figure 3.8.7(d)) opens. Type the data shown in the windows under "Add Bound" and press first, the Add button to the right, and then press the Done button at the bottom.
Once the bounds on x are known to Maple, choose "Maximize" in the Optimization Assistant, and press the Solve button to obtain the maximum of Ax, and the corresponding value of x where this maximum occurs. For the sake of completeness, calculate the corresponding value of y from the equation 5 x2+12 y2=60.
Analytic Solution
Define the objective function Ax
Write Ax=… Context Panel: Assign Function
Ax=2 x 51−x2/12→assign as functionA
Find the critical number and the corresponding maximum
Write A′x=0 and press the Enter key.
Context Panel: Solve≻Solve
A′x=0
13⁢−15⁢x2+180−5⁢x2−15⁢x2+180=0
→solve
x=−6,x=6
Write the positive root. Context Panel: Approximate≻10 (digits)
6→at 10 digits2.449489743
Evaluate A at the critical number. Context Panel: Evaluate and Display Inline Context Panel: Simplify≻Simplify
A6 = 13⁢6⁢90= simplify 2⁢3⁢5
Evaluate A at the critical number. Context Panel: Approximate≻10 (digits)
A6→at 10 digits7.745966693
Expression palette: Evaluation template Evaluate constraint ellipse at x=6 Press the Enter key.
5 x2+12 y2=60x=a|f(x)x=6
12⁢y2+30=60
y=12⁢10,y=−12⁢10
Write the positive solution. Context Panel: Approximate≻10 (digits)
10/2→at 10 digits1.581138830
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