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Worksheet · Calculus I · Unit 2B

Calculus Optimization Worksheet with Complete Solutions

Sixteen substantial modeling problems with constraints, domains, critical points, and verification.

16 problems90 min estimated timeIntermediate to advanced progression

What is included

Practice open-box, fencing, geometry, distance, cost, and revenue optimization with complete derivations.

Skills assessed

  • constraint construction
  • domain reasoning
  • extrema verification

Prerequisites

  • derivative rules
  • critical points
Optimization Worksheet instructional sequence
Sixteen substantial modeling problems with constraints, domains, critical points, and verification. The numbered labels and written sequence preserve meaning without relying on color.
Long description

Read the diagram from top to bottom. Each numbered box names one decision or mathematical operation. Arrows show the required order; the text labels remain the complete interpretation in print, dark mode, and nonvisual reading.

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  1. Squares of side x are cut from a 12 by 12 sheet. Maximize the open box volume.
  2. Squares of side x are cut from a 20 by 12 sheet. Maximize volume.
  3. Use 240 m of fence for three sides of a riverside rectangle. Maximize area.
  4. A 600 m rectangle has one divider parallel to its width. Maximize area.
  5. A poster has area 384 cm² with 2 cm side margins and 3 cm top and bottom margins. Minimize total paper area.
  6. Find the closed cylinder of volume 250\pi with minimum surface area.
  7. Find the point on y=x² closest to (0,3).
  8. Maximize the area of a rectangle inscribed under y=\sqrt{25-x^2}.
  9. Cut 20 m of wire into a square and a circle to minimize total area.
  10. Demand is p=80-2q. Maximize revenue.
  11. Minimize average cost C(q)/q when C(q)=q²+100q+2500.
  12. A 10 ft ladder rests against a wall. Maximize the area of the right triangle it forms.
  13. A cone has volume 72\pi. Under the material model S=\pi r²+\pi rh, minimize S.
  14. A Norman window has perimeter 20. Maximize area.
  15. A boat 3 km offshore must reach a town 8 km downshore. Row at 3 km/h and walk at 5 km/h. Minimize time.
  16. Maximize f(x)=x(6-x) on the closed interval [1,5].

Complete worked solutions

Every problem has a source-matched answer and independently reviewed derivation.

01

Problem 1: Squares of side x are cut from a 12 by 12 sheet. Maximize the open box volume.

Answer: x=2, Vmax⁡=128x=2,\ V_{\max}=128

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. V=x(12-2x)^2 on 0<x<6; V'=12(x-2)(x-6), so the interior maximum is x=2.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is x=2, Vmax⁡=128x=2,\ V_{\max}=128.
02

Problem 2: Squares of side x are cut from a 20 by 12 sheet. Maximize volume.

Answer: x=16−763≈2.43x=\frac{16-\sqrt{76}}3\approx2.43

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. V=x(20-2x)(12-2x). Solve V'=12x^2-128x+240=0 and retain the feasible critical point.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is x=16−763≈2.43x=\frac{16-\sqrt{76}}3\approx2.43.
03

Problem 3: Use 240 m of fence for three sides of a riverside rectangle. Maximize area.

Answer: x=60, y=120, A=7200x=60,\ y=120,\ A=7200

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. With 2x+y=240, A=x(240-2x); A'=240-4x=0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is x=60, y=120, A=7200x=60,\ y=120,\ A=7200.
04

Problem 4: A 600 m rectangle has one divider parallel to its width. Maximize area.

Answer: w=100,ℓ=150, A=15000w=100,\ell=150,\ A=15000

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. The constraint is 3w+2\ell=600. Substitute \ell=(600-3w)/2 into A=w\ell.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is w=100,ℓ=150, A=15000w=100,\ell=150,\ A=15000.
05

Problem 5: A poster has area 384 cm² with 2 cm side margins and 3 cm top and bottom margins. Minimize total paper area.

Answer: printed width=16, printed height=24printed\ width=16,\ printed\ height=24

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. For printed width x, paper area is (x+4)(384/x+6). Set its derivative to zero.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is printed width=16, printed height=24printed\ width=16,\ printed\ height=24.
06

Problem 6: Find the closed cylinder of volume 250\pi with minimum surface area.

Answer: r=5, h=10r=5,\ h=10

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Use h=250/r^2 in S=2\pi r^2+2\pi rh; solve S'=0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is r=5, h=10r=5,\ h=10.
07

Problem 7: Find the point on y=x² closest to (0,3).

Answer: (±5/2,5/2)(\pm\sqrt{5/2},5/2)

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Minimize D^2=x^2+(x^2-3)^2. Critical points satisfy 2x(2x^2-5)=0; compare values.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is (±5/2,5/2)(\pm\sqrt{5/2},5/2).
08

Problem 8: Maximize the area of a rectangle inscribed under y=\sqrt{25-x^2}.

Answer: width=52, height=5/2, A=25width=5\sqrt2,\ height=5/\sqrt2,\ A=25

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. A=2x\sqrt{25-x^2}; maximize A²=4x²(25-x²).
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is width=52, height=5/2, A=25width=5\sqrt2,\ height=5/\sqrt2,\ A=25.
09

Problem 9: Cut 20 m of wire into a square and a circle to minimize total area.

Answer: square length=804+π, circle length=20π4+πsquare\ length=\frac{80}{4+\pi},\ circle\ length=\frac{20\pi}{4+\pi}

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Let x go to the square: A=x²/16+(20-x)²/(4\pi). Solve A'=0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is square length=804+π, circle length=20π4+πsquare\ length=\frac{80}{4+\pi},\ circle\ length=\frac{20\pi}{4+\pi}.
10

Problem 10: Demand is p=80-2q. Maximize revenue.

Answer: q=20, p=40, R=800q=20,\ p=40,\ R=800

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. R=q(80-2q); R'=80-4q=0 and R''<0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is q=20, p=40, R=800q=20,\ p=40,\ R=800.
11

Problem 11: Minimize average cost C(q)/q when C(q)=q²+100q+2500.

Answer: q=50q=50

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Average cost is q+100+2500/q. Set 1-2500/q²=0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is q=50q=50.
12

Problem 12: A 10 ft ladder rests against a wall. Maximize the area of the right triangle it forms.

Answer: x=y=52, A=25x=y=5\sqrt2,\ A=25

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. With x²+y²=100, maximize A=xy/2; symmetry or differentiation gives x=y.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is x=y=52, A=25x=y=5\sqrt2,\ A=25.
13

Problem 13: A cone has volume 72\pi. Under the material model S=\pi r²+\pi rh, minimize S.

Answer: r=1083=343, h=21083=643r=\sqrt[3]{108}=3\sqrt[3]{4},\ h=2\sqrt[3]{108}=6\sqrt[3]{4}

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Use h=216/r², so S=\pi r²+216\pi/r. Then S'=2\pi r-216\pi/r²=0 gives r³=108 and h=216/r²=2r.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is r=1083=343, h=21083=643r=\sqrt[3]{108}=3\sqrt[3]{4},\ h=2\sqrt[3]{108}=6\sqrt[3]{4}.
14

Problem 14: A Norman window has perimeter 20. Maximize area.

Answer: r=204+πr=\frac{20}{4+\pi}

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Use width 2r and rectangular height h. The perimeter gives h=(20-(2+\pi)r)/2; maximize 2rh+\pi r²/2.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is r=204+πr=\frac{20}{4+\pi}.
15

Problem 15: A boat 3 km offshore must reach a town 8 km downshore. Row at 3 km/h and walk at 5 km/h. Minimize time.

Answer: land about 5.75 km before townland\ about\ 5.75\ km\ before\ town

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Let x be the downshore rowing distance. T=\sqrt{x²+9}/3+(8-x)/5; solve T'=0.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is land about 5.75 km before townland\ about\ 5.75\ km\ before\ town.
16

Problem 16: Maximize f(x)=x(6-x) on the closed interval [1,5].

Answer: x=3, f(3)=9x=3,\ f(3)=9

Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.

  1. Check the critical point x=3 and both endpoints; the largest listed value is 9.
  2. Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
  3. Therefore the result is x=3, f(3)=9x=3,\ f(3)=9.

Common errors

  • Differentiating before reducing to one variable.
  • Ignoring the feasible domain.
  • Reporting a critical point without verifying the optimum.

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