Calculus Optimization Worksheet with Complete Solutions
Sixteen substantial modeling problems with constraints, domains, critical points, and verification.
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
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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- Squares of side x are cut from a 12 by 12 sheet. Maximize the open box volume.
- Squares of side x are cut from a 20 by 12 sheet. Maximize volume.
- Use 240 m of fence for three sides of a riverside rectangle. Maximize area.
- A 600 m rectangle has one divider parallel to its width. Maximize area.
- A poster has area 384 cm² with 2 cm side margins and 3 cm top and bottom margins. Minimize total paper area.
- Find the closed cylinder of volume 250\pi with minimum surface area.
- Find the point on y=x² closest to (0,3).
- Maximize the area of a rectangle inscribed under y=\sqrt{25-x^2}.
- Cut 20 m of wire into a square and a circle to minimize total area.
- Demand is p=80-2q. Maximize revenue.
- Minimize average cost C(q)/q when C(q)=q²+100q+2500.
- A 10 ft ladder rests against a wall. Maximize the area of the right triangle it forms.
- A cone has volume 72\pi. Minimize slant-independent material S=\pi r²+\pi rh.
- A Norman window has perimeter 20. Maximize area.
- 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.
- 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.
Problem 1: Squares of side x are cut from a 12 by 12 sheet. Maximize the open box volume.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- V=x(12-2x)^2 on 0<x<6; V'=12(x-2)(x-6), so the interior maximum is x=2.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 2: Squares of side x are cut from a 20 by 12 sheet. Maximize volume.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- V=x(20-2x)(12-2x). Solve V'=12x^2-128x+240=0 and retain the feasible critical point.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 3: Use 240 m of fence for three sides of a riverside rectangle. Maximize area.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- With 2x+y=240, A=x(240-2x); A'=240-4x=0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 4: A 600 m rectangle has one divider parallel to its width. Maximize area.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- The constraint is 3w+2\ell=600. Substitute \ell=(600-3w)/2 into A=w\ell.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
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:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- For printed width x, paper area is (x+4)(384/x+6). Set its derivative to zero.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 6: Find the closed cylinder of volume 250\pi with minimum surface area.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Use h=250/r^2 in S=2\pi r^2+2\pi rh; solve S'=0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 7: Find the point on y=x² closest to (0,3).
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Minimize D^2=x^2+(x^2-3)^2. Critical points satisfy 2x(2x^2-5)=0; compare values.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 8: Maximize the area of a rectangle inscribed under y=\sqrt{25-x^2}.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- A=2x\sqrt{25-x^2}; maximize A²=4x²(25-x²).
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 9: Cut 20 m of wire into a square and a circle to minimize total area.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Let x go to the square: A=x²/16+(20-x)²/(4\pi). Solve A'=0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 10: Demand is p=80-2q. Maximize revenue.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- R=q(80-2q); R'=80-4q=0 and R''<0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 11: Minimize average cost C(q)/q when C(q)=q²+100q+2500.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Average cost is q+100+2500/q. Set 1-2500/q²=0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 12: A 10 ft ladder rests against a wall. Maximize the area of the right triangle it forms.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- With x²+y²=100, maximize A=xy/2; symmetry or differentiation gives x=y.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 13: A cone has volume 72\pi. Minimize slant-independent material S=\pi r²+\pi rh.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Use h=216/r², then minimize S=\pi r²+216\pi/r.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 14: A Norman window has perimeter 20. Maximize area.
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Use width 2r and rectangular height h. The perimeter gives h=(20-(2+\pi)r)/2; maximize 2rh+\pi r²/2.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
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:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Let x be the downshore rowing distance. T=\sqrt{x²+9}/3+(8-x)/5; solve T'=0.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Problem 16: Maximize f(x)=x(6-x) on the closed interval [1,5].
Answer:
Why this method: Constrained one-variable optimization matches the mathematical structure before any algebraic cleanup.
- Check the critical point x=3 and both endpoints; the largest listed value is 9.
- Check feasibility, endpoints, and the sign or second derivative before declaring the optimum.
Common errors
- Differentiating before reducing to one variable.
- Ignoring the feasible domain.
- Reporting a critical point without verifying the optimum.