Chapter 3: Applications of Differentiation
Section 3.8: Optimization
Example 3.8.11
A can in the shape of a right-circular cylinder, fashioned from a rectangle rolled into a cylinder and disks welded at the top and bottom, must have volume V. What are the dimensions of the least-cost can if the side costs a per square inch and the top and bottom cost b per square inch?
Solution
Analysis
Figure 3.8.11(a) animates rolling a rectangular sheet of material into a cylinder.
If the radius of the resulting cylinder is r, then the circumference (and hence, the width of the sheet) is 2 π r.
Of course, the height of the cylinder is the height of the rectangle, say, h.
The volume of the cylinder is then V=π r2h, which is a constraint on the objective function, itself the total cost of construction.
use plots, plottools in ex3811:=module() export F; F:=proc(a) local p1,p2; p1:=plottools[polygon]([[cos(a),sin(a),0],[cos(a)-(2*Pi-a)*sin(a),sin(a)+(2*Pi-a)*cos(a),0],[cos(a)-(2*Pi-a)*sin(a),sin(a)+(2*Pi-a)*cos(a),3],[cos(a),sin(a),3]],color=green); p2:=plot3d(1,theta=0..a,z=0..3,coords=cylindrical,color=red): display(p1,p2,scaling=constrained); end proc: end module: end use: plots:-animate(ex3811:-F,[a],a=0..2*Pi,paraminfo=false,orientation=[-160,75,0],axes=none,lightmodel=none);
Figure 3.8.11(a) Animation: rectangle to cylinder
The objective function is the total cost, 2 π r h a+2π r2 b.
Solve the constraint equation for h=Vπ r2 so that the cost function is then 2 π r Vπ r2 a+2 π r2 b or
Cr=2 a Vr+2 π r2 b
Analytic Solution
Initialize and define the cost function Cr
Tools≻Load Package: Real Domain
Loading RealDomain
Control-drag Cr=… Context Panel: Assign Function
Cr=2 a Vr+2 π r2 b→assign as functionC
Obtain the critical number
Write C′r=0 and press the Enter key.
Context Panel: Solve≻Obtain Solutions for≻r
Context Panel: Assign to a Name≻Rmin
C′r=0
−2⁢a⁢Vr2+4⁢π⁢r⁢b=0
→solutions for r
12⁢22/3⁢a⁢V⁢b21/3π1/3⁢b
→assign to a name
Rmin
Unload the Real Domain package
Tools≻Unload Package: Real Domain
Unloading RealDomain
Apply the Second-Derivative test
Write C″r Context Panel: Evaluate and Display Inline Context Panel: Evaluate at a Point≻r=Rmin (C″Rmin>0 ⇒ Rmin is a relative minimum)
C″r = 4⁢a⁢Vr3+4⁢π⁢b→evaluate at point12⁢π⁢b
Obtain the minimum cost
Write CRmin and press the Enter key.
Context Panel: Simplify≻Assuming Positive
CRmin
2⁢a⁢V⁢π1/3⁢b⁢21/3a⁢V⁢b21/3+π1/3⁢21/3⁢a⁢V⁢b22/3b
→assuming positive
3⁢a2/3⁢V2/3⁢π1/3⁢b1/3⁢21/3
Define λ=a/b so that the dimensions minimizing cost are
Rmin = 12⁢22/3⁢a⁢V⁢b21/3π1/3⁢b=12⁢22/3⁢a1/3⁢V1/3π1/3⁢b1/3 = V2 π1/3 ab1/3=V2 π1/3λ1/3
and
Vπ Rmin2 = V⁢b2⁢22/3π1/3⁢a⁢V⁢b22/3 = =V1/3⁢b2/3⁢22/3π1/3⁢a2/3 = 4π1/3V1/3λ2/3
Figure 3.8.11(b) shows how the shape of a can with V=1 varies as the relative costs a/b=λ vary. For λ<0.25, the can looks like a mailing tube, so for the sake of the image all such values of λ are set to 0.25.
λ= =
Figure 3.8.11(b) How relative cost affects shape of can
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