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Applications of Differentiation

3.7 Optimization Problems

Focus on a one-variable objective, open-box optimization, constrained versus unconstrained maxima.

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Ideas in this lesson

01A one-variable objectiveWorked steps · five questions02Open-box optimizationWorked steps · five questions03Constrained versus unconstrained maximaWorked steps · five questions
Idea 1 of 3

A one-variable objective

Use the constraint to express the quantity being optimized in one variable. State its feasible interval.

Worked example 1

A rectangle has perimeter 88. Find the side length that maximizes area.
  1. y=4−xy = 4 - x; A(x)=x(4−x)A(x) = x(4 - x)
  2. A′(x)=4−2x=0  ⟹  x=2A'(x) = 4 - 2x = 0 \implies x = 2; A′′=−2A'' = -2
  3. Maximizing side =2= 2

Answer: 22

Worked example 2

A rectangle has perimeter 1212. Find the maximum area.
  1. y=6−xy = 6 - x; A(x)=x(6−x)A(x) = x(6 - x)
  2. A′(x)=6−2x=0  ⟹  x=3A'(x) = 6 - 2x = 0 \implies x = 3; A′′=−2A'' = -2
  3. A(3)=32=9A(3) = 3^{2} = 9

Answer: 99

Idea 2 of 3

Open-box optimization

The cut size changes both height and base dimensions. Reject roots outside the geometric domain.

Worked example 1

Squares of side xx are cut from a 1212 by 1212 sheet to form an open box. Find the volume at x=2x = 2.
  1. V(x)=x(12−2x)2V(x) = x(12 - 2x)^{2}; 0<x<60 < x < 6
  2. V′(x)=(12−2x)(12−6x)=0  ⟹  x=2V'(x) = (12 - 2x)(12 - 6x) = 0 \implies x = 2
  3. V(0)=V(6)=0V(0) = V(6) = 0; V(2)=128V(2) = 128
  4. Requested value =128= 128

Answer: 128128

Worked example 2

Squares of side xx are cut from a 1818 by 1818 sheet to form an open box. Find the positive interior cut that maximizes volume.
  1. V(x)=x(18−2x)2V(x) = x(18 - 2x)^{2}; 0<x<90 < x < 9
  2. V′(x)=(18−2x)(18−6x)=0  ⟹  x=3V'(x) = (18 - 2x)(18 - 6x) = 0 \implies x = 3
  3. V(0)=V(9)=0V(0) = V(9) = 0; V(3)=432V(3) = 432
  4. Requested value =3= 3

Answer: 33

Idea 3 of 3

Constrained versus unconstrained maxima

An interior critical point may be infeasible. Check boundary values as well as valid critical points.

Worked example 1

A rectangle has perimeter 88. Find the side length that maximizes area.
  1. y=4−xy = 4 - x; A(x)=x(4−x)A(x) = x(4 - x)
  2. A′(x)=4−2x=0  ⟹  x=2A'(x) = 4 - 2x = 0 \implies x = 2; A′′=−2A'' = -2
  3. Maximizing side =2= 2

Answer: 22

Worked example 2

A rectangle has perimeter 1212. Find the maximum area.
  1. y=6−xy = 6 - x; A(x)=x(6−x)A(x) = x(6 - x)
  2. A′(x)=6−2x=0  ⟹  x=3A'(x) = 6 - 2x = 0 \implies x = 3; A′′=−2A'' = -2
  3. A(3)=32=9A(3) = 3^{2} = 9

Answer: 99

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.7,p3.7, p. 215215. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.