Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.4
A wire of length L (cm) is cut into two pieces. One piece is bent into an equilateral triangle; the other, into a circle. Find the maximum and minimum values for the sum of the areas enclosed by the two pieces of wire.
Solution
Analysis
Cut the wire into two lengths, x and L−x. Fashion the triangle from the piece of length x, and the circle from the piece of length L−x.
Figure 3.8.4(a) shows the triangle, the circle, and a graph of the combined area when L=10, wherein the length x is determined by the position of the accompanying slider.
The green dot, synchronized to the slider, traverses the graph of Ax, the combined areas of the triangle and circle.
From Figure 3.8.4(a), deduce that there is a minimum near x=6 and a maximum at the left endpoint where x=0, that is, when the complete wire is used to fashion the circle.
x= = ⇒ A=
Figure 3.8.4(a) Area of triangle and circle
Taking the area of a triangle as 12a b sinC, the area of the equilateral triangle is
12x3⋅x3⋅32=x2336
The circumference of the circle is 2 π r=L−x, so the radius is r=L−x2 π, and the area is
π r2=π L−x2 π2=L−x24 π
The combined area, the objective function to be minimized, is Ax=x2336+L−x24 π.
The implied constraints are 0≤x≤L.
Analytic Solution
Define the objective function Ax
Control-drag Ax=… Context Panel: Assign Function
Ax=x2336+L−x24 π→assign as functionA
Obtain the critical number
Write the equation for the critical number; Press the enter key.
Context Panel: Solve≻Obtain Solutions for≻x
Context Panel: Assign to a Name≻c
A′x=0
118x3−12L−xπ=0
→solutions for x
9Lπ3+9
→assign to a name
c
Implement the Second-Derivative test
Find A″c>0, so c,Ac is a minimum.
A″ = 1183+12π
Calculate Ac
Ac = 94L23π3+92+14L−9Lπ3+92π= simplify 14L23π3+9
Calculate A0
From Figure 3.8.5, the absolute maximum occurs when x=0.
A0 = 14L2π
Sample these calculations for L=10
Expression palette: Evaluation template Evaluate A0 with L=10
Context Panel: Approximate≻5 (digits)
A0x=a|f(x)L=10 = 25π→at 5 digits7.9578
Expression palette: Evaluation template Evaluate c with L=10
cx=a|f(x)L=10 = 90π3+9→at 5 digits6.2318
Expression palette: Evaluation template Evaluate Ac with L=10
Context Panel: Simplify≻Simplify
Acx=a|f(x)L=10 = 2253π3+92+1410−90π3+92π= simplify 253π3+9→at 5 digits2.9982
The graphical results from Figure 3.8.4(a) agree with the numeric results calculated for the analytic solution when L=10.
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