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Introduction to Optimization Problems - AP Calculus BC

  1. A rectangle with perimeter 40 has length ll and width ww. To find the rectangle with maximum area, what setup is correct?
    Answer: Maximize A=lwA=lw subject to 2l+2w=402l+2w=40 This problem involves maximizing the area of a rectangle with fixed perimeter constraint. The rectangle has length l and width w with perimeter constraint 2l + 2w = 40, which simplifies to l + w = 20. We want to maximize area A = lw subject to this constraint. Using w = 20 - l, we get A(l) = l(20 - l) = 20l - l². This is maximized when l = 10, giving w = 10 and maximum area 100. Choice B incorrectly minimizes area, but we want the largest possible rectangle. The key insight is that for fixed perimeter, the rectangle with maximum area is always a square (l = w).
  2. A rectangular field is divided into two equal rectangles by a fence parallel to one side. If total fencing is fixed, what should be minimized?
    Answer: Minimize total fencing length as a function of the field's side lengths for a fixed area This problem requires minimizing the total fencing needed for a field with internal division. A rectangular field is divided into two equal rectangles by an internal fence parallel to one side. If the field has length l and width w, the total fencing includes the perimeter (2l + 2w) plus the internal divider (length w), giving total fencing 2l + 3w or 2l + 2w + l depending on orientation. Since total fencing is fixed, we want to minimize fencing for a given area, or equivalently, maximize area for given fencing. Choice A correctly identifies minimizing fencing for fixed area. The key insight is that internal divisions require additional fencing beyond the basic perimeter.
  3. A cable is to be strung from point A to B with a support pole at variable point P. If cost depends on total cable length, what is minimized?
    Answer: The total cable length as a function of the location of point P This problem involves minimizing the total length of cable required to connect two points via an intermediate support point. Points A and B are fixed, and point P is variable along some path (line, curve, or region). The total cable length is the sum of distances AP + PB, which depends on the location of P. If cable cost is proportional to length, then minimizing cost is equivalent to minimizing total cable length as a function of P's position. Choice B incorrectly treats distance from A to P as a function of cable cost, but cost depends on length, not vice versa. The modeling strategy is to express total cost as the sum of segment costs, with each segment cost depending on geometric distances.
  4. A rectangle is to be placed inside a circle of radius 5. To maximize area, what should be optimized?
    Answer: Maximize A=lwA=lw subject to the constraint that the diagonal satisfies l2+w2=100l^2+w^2=100 This problem involves maximizing the area of a rectangle inscribed in a circle. For a rectangle inscribed in a circle of radius 5, the diagonal of the rectangle equals the diameter of the circle, which is 10. If the rectangle has length l and width w, then l² + w² = 100 (since diagonal = 10). We want to maximize the area A = lw subject to this constraint. Choice B incorrectly minimizes area, but we want the largest possible rectangle. Using the method of Lagrange multipliers or substitution, the maximum occurs when l = w = 5√2, giving maximum area 50. The approach is to use the geometric constraint (rectangle inscribed in circle means diagonal equals diameter) to set up the optimization problem.
  5. A company sells xx items; profit is revenue minus cost. If R(x)=50xR(x)=50x and C(x)=200+10xC(x)=200+10x, what should be maximized?
    Answer: Maximize profit P(x)=R(x)−C(x)=40x−200P(x)=R(x)-C(x)=40x-200 This problem requires maximizing profit, which is the difference between revenue and cost functions. Given revenue R(x) = 50x and cost C(x) = 200 + 10x, the profit function is P(x) = R(x) - C(x) = 50x - (200 + 10x) = 40x - 200. Since profit represents the company's financial objective, we maximize P(x) = 40x - 200. Choice A incorrectly suggests maximizing revenue alone, but this ignores costs entirely. The key insight is that business optimization focuses on profit (net benefit) rather than gross measures like revenue, requiring the objective function to account for both benefits and costs.
  6. A triangle has base bb on the xx-axis and vertex on the line y=8y=8. If area is to be maximized with fixed equal sides, what should be maximized?
    Answer: Maximize the triangle's area as a function of the base endpoints subject to the side-length constraint This problem requires maximizing the area of a triangle with specific constraints on vertex location and side lengths. The triangle has base b on the x-axis and vertex on the line y = 8. If the triangle has two equal sides (isosceles), then the vertex is directly above the midpoint of the base, at coordinates (b/2, 8). The triangle's area is A = (1/2) × base × height = (1/2) × b × 8 = 4b. However, there's also a constraint that the sides are fixed equal lengths. If the equal sides have length s, then by Pythagorean theorem: s² = (b/2)² + 8² = b²/4 + 64. For fixed s, this constrains b, making this an area maximization subject to the side-length constraint. Choice A correctly identifies this setup. The approach is to express triangle area in terms of the base, then apply the side-length constraint to determine the feasible values.
  7. A company's demand is p=80−2xp=80-2x dollars per unit. Revenue is R=xpR=xp. What should be maximized to find best sales level?
    Answer: Maximize revenue R(x)=x(80−2x)R(x)=x(80-2x) This problem involves maximizing revenue using a linear demand function. The demand relationship is p = 80 - 2x, where p is price and x is quantity. Revenue is R = xp = x(80 - 2x) = 80x - 2x². This is a quadratic function in x, and we maximize it to find the optimal sales level. The maximum occurs at x = 80/(2·2) = 20, giving maximum revenue R = 80(20) - 2(20)² = 1600 - 800 = 800. Choice B incorrectly minimizes revenue, which would minimize profit. The fundamental approach is to substitute the demand function into the revenue equation and optimize the resulting quadratic function.
  8. A student wants to minimize the distance from point (2,5)(2,5) to a point (x,0)(x,0) on the xx-axis. What should be minimized?
    Answer: Minimize D(x)=(x−2)2+25D(x)=\sqrt{(x-2)^2+25} for all real xx This problem requires minimizing the distance from a fixed point to a variable point constrained to lie on the x-axis. The fixed point is (2,5) and the variable point is (x,0). The distance is D(x) = √[(x-2)² + (0-5)²] = √[(x-2)² + 25]. Since we want to minimize this distance, and x can be any real number, we minimize D(x) = √[(x-2)² + 25]. Choice C incorrectly uses (x+2)² instead of (x-2)², and choice D unnecessarily restricts the domain. The fundamental approach is to use the distance formula and recognize that minimizing distance is equivalent to minimizing the squared distance when the square root complicates differentiation.
  9. A poster has area 384 cm2384\text{ cm}^2 with 2 cm margins on sides and 3 cm margins top and bottom. What should be maximized?
    Answer: The printed (non-margin) area as a function of the poster's width and height This problem requires maximizing the useful printed area within a fixed total poster area constraint. The poster has total area 384 cm², with margins of 2 cm on sides and 3 cm on top and bottom. If the poster has width w and height h, then wh = 384. The printed area has width (w - 4) and height (h - 6), so the printed area is (w - 4)(h - 6). We want to maximize this printed area as a function of the poster's dimensions. Choice A incorrectly focuses on total area, but that's fixed at 384 cm². The strategy is to identify what quantity provides value (printed area) and optimize that subject to the physical constraints.
  10. A cylindrical silo must have height 3 times its radius and volume 100π100\pi. What should be minimized to reduce exterior paint?
    Answer: Minimize lateral surface area S=2πrhS=2\pi rh using h=3rh=3r and πr2h=100π\pi r^2 h=100\pi This problem requires minimizing the lateral surface area of a cylinder with geometric and volume constraints. The silo has height h=3rh = 3r and volume 100π100\pi. Using the volume constraint πr2h=πr2(3r)=3πr3=100π\pi r^2 h = \pi r^2 (3r) = 3\pi r^3 = 100\pi, we get r3=100/3r^3 = 100/3, so r=100/33r = \sqrt[3]{100/3}. The lateral surface area is S=2πrh=2πr(3r)=6πr2S = 2\pi r h = 2\pi r (3r) = 6\pi r^2. Since we want to minimize exterior paint (lateral surface area), we minimize SS with respect to rr using the given constraints. Choice B incorrectly maximizes surface area, increasing paint usage. The key insight is to use both geometric constraints (h=3rh = 3r) and performance constraints (volume = 100π100\pi) to express surface area as a function that can be optimized.
  11. A cone-shaped paper cup must hold volume VV with fixed slant height ss. To maximize volume, what variable should be optimized?
    Answer: Maximize V(r)=13πr2hV(r)=\tfrac13\pi r^2h expressed in terms of rr using r2+h2=s2r^2+h^2=s^2 This problem requires maximizing the volume of a cone with fixed slant height constraint. For a cone with base radius rr, height hh, and slant height ss, the relationship is r2+h2=s2r^2 + h^2 = s^2. The volume is V=(1/3)πr2hV = (1/3)\pi r^2 h. With ss fixed, we can write h=s2−r2h = \sqrt{s^2 - r^2} and substitute to get V(r)=(1/3)πr2s2−r2V(r) = (1/3)\pi r^2 \sqrt{s^2 - r^2}. We maximize this volume as a function of rr, with domain 0≤r≤s0 \leq r \leq s. Choice B incorrectly minimizes volume, but we want the largest possible cup. The key insight is to use the geometric constraint (relating rr, hh, and ss) to express volume as a function of one variable, then optimize that single-variable function.
  12. A rectangle has its upper corners on the line y=10−xy=10-x in the first quadrant and sides on axes. What should be maximized?
    Answer: Maximize area A(x)=x(10−x)A(x)=x(10-x) for 0≤x≤100\le x\le 10 This problem requires maximizing the area of a rectangle with vertices constrained by a line and the axes. The rectangle has one side on the x-axis from (0,0) to (x,0) and extends up to the line y = 10 - x. The rectangle's width is x and height is 10 - x, giving area A(x) = x(10 - x) = 10x - x². The domain is 0 ≤ x ≤ 10 since both width and height must be non-negative. Choice B incorrectly minimizes area, but we want maximum enclosed space. The key insight is to parameterize the rectangle using one coordinate, then use the constraint (the line equation) to express the area as a single-variable function.
  13. A rectangular aquarium must have volume 2000 L with length twice the width. If glass area is to be minimized, what is optimized?
    Answer: Minimize total glass surface area as a function of width using l=2wl=2w and lwh=2000lwh=2000 This problem requires minimizing the glass surface area for an aquarium with volume and geometric constraints. The aquarium has volume 2000 L with length l = 2w (twice the width). The volume constraint is lwh = 2wh = 2000, so wh = 1000. The glass surface area includes bottom (2w²), front and back (2·2w·h = 4wh), and two ends (2·w·h = 2wh), giving total area S = 2w² + 4wh + 2wh = 2w² + 6wh = 2w² + 6(1000) = 2w² + 6000. Wait, that's not right. Using wh = 1000, we get S = 2w² + 6wh = 2w² + 6000/w? Let me reconsider: S = 2w² + 4wh + 2wh = 2w² + 6wh, and using h = 1000/w gives S(w) = 2w² + 6000/w. Choice A correctly identifies minimizing glass surface area. The approach is to use the volume constraint to eliminate one variable from the surface area expression.
  14. A rectangular prism has a square base and must have total surface area 150. To maximize volume, what should be maximized?
    Answer: Maximize volume V=x2hV=x^2h subject to 2x2+4xh=1502x^2+4xh=150 This problem involves maximizing volume for a rectangular prism with square base and fixed surface area constraint. The prism has square base with side x and height h, with total surface area constraint 2x2+4xh=1502x^2 + 4xh = 150 (including base, top, and four sides). We want to maximize volume V=x2hV = x^2 h subject to this surface area constraint. Using the constraint to solve for h: h=150−2x24xh = \frac{150 - 2x^2}{4x}, so V(x)=x2[150−2x24x]=x(150−2x2)4V(x) = x^2 \left[ \frac{150 - 2x^2}{4x} \right] = \frac{x(150 - 2x^2)}{4}. Choice B incorrectly minimizes volume, but we want maximum storage capacity. The approach is to use the surface area constraint to eliminate one variable from the volume expression, then optimize the resulting function.
  15. A rectangular storage bin has a square base and must fit under a shelf 1.5 m high. If volume is to be maximized, what is optimized?
    Answer: Maximize volume V=x2hV=x^2h with constraint 0<h≤1.50<h\le 1.5 This problem requires maximizing the volume of a storage bin with a square base and height constraint. The bin has square base with side x and height h, with constraint h ≤ 1.5 m (to fit under the shelf). To maximize storage capacity, we want to maximize volume V = x²h. Since there's no other constraint given, the maximum occurs when h takes its maximum allowed value h = 1.5. Thus we maximize V = 1.5x² by choosing x as large as practical considerations allow. Choice B incorrectly minimizes volume, reducing storage capacity. The key insight is that when one dimension is bounded by a constraint, maximizing volume often involves setting that dimension to its maximum allowed value.
  16. A square and a circle have total area 100. If the sum of perimeters is to be minimized, what should be minimized?
    Answer: Minimize P=4s+2πrP=4s+2\pi r subject to s2+πr2=100s^2+\pi r^2=100 This problem involves minimizing the total perimeter of a square and circle with fixed combined area. Let the square have side length s and the circle have radius r, with total area constraint s2+πr2=100s^2 + \pi r^2 = 100. The total perimeter is P=4s+2πrP = 4s + 2\pi r. We want to minimize this perimeter subject to the area constraint. Using the constraint to eliminate one variable, say s=100−πr2s = \sqrt{100 - \pi r^2}, we get P(r)=4100−πr2+2πrP(r) = 4\sqrt{100 - \pi r^2} + 2\pi r. Choice B incorrectly maximizes perimeter, but we want to minimize the total boundary length. The modeling approach is to set up the perimeter minimization with the area constraint, which determines the optimal allocation of area between the two shapes.
  17. A poster has 300 cm2^2 of printed area and 2 cm margins on all sides; what objective should be minimized?​
    Answer: The total poster area (w+4)(h+4)(w+4)(h+4) subject to wh=300wh=300 This poster optimization problem requires identifying what to minimize given a constraint. The printed area is fixed at wh = 300 cm², and with 2 cm margins on all sides, the total poster dimensions are (w+4) by (h+4). To minimize paper usage, we minimize the total poster area (w+4)(h+4) subject to the constraint wh = 300. The printed area (choice A) cannot be minimized as it's already fixed at 300 cm². When dealing with margin problems, optimize the total dimensions while keeping the inner content area fixed.
  18. A farmer has 200 m of fencing to enclose a rectangular pen; one side borders a river needing no fence. What should be maximized?
    Answer: The area of the rectangular pen as a function of the two fenced side lengths This problem requires setting up an optimization to maximize enclosed area with a constraint on available fencing. The farmer wants to create the largest possible rectangular pen using 200 m of fencing, where one side borders a river and needs no fence. If the pen has width w (perpendicular to river) and length l (parallel to river), then the constraint is w + w + l = 200, or 2w + l = 200. The objective is to maximize the area A = wl as a function of the fence dimensions. Choice B incorrectly suggests maximizing perimeter, but perimeter is fixed by the constraint. The key insight is that optimization problems maximize or minimize an objective function (here, area) subject to constraint equations involving the decision variables.
  19. A rectangular picture frame has outer dimensions xx by yy with fixed outer perimeter 100. Frame border width is constant. What should be maximized?
    Answer: Maximize the visible picture area as a function of xx and yy subject to 2x+2y=1002x+2y=100 This problem involves maximizing the visible picture area in a frame with fixed outer perimeter and constant border width. The frame has outer dimensions x by y with constraint 2x + 2y = 100 (fixed outer perimeter). If the border has constant width w on all sides, the visible picture area is (x-2w)(y-2w). Since w is constant and the outer perimeter is fixed, we want to maximize the visible area as a function of x and y subject to the perimeter constraint. Using y = 50 - x, the visible area becomes A(x) = (x-2w)(50-x-2w) = (x-2w)(50-x-2w). Choice B incorrectly minimizes visible area. The modeling strategy is to account for the border width when expressing the visible area, then optimize subject to the outer dimension constraints.
  20. A tank's cross-section is a rectangle of width ww and height hh with fixed area wh=36wh=36. If material for the sides is proportional to perimeter, what is minimized?
    Answer: Minimize P(w)=2w+2hP(w)=2w+2h subject to wh=36wh=36 This problem requires minimizing the perimeter of a rectangle with fixed area to reduce material costs. The rectangular cross-section has width w and height h with fixed area wh = 36. The perimeter is P = 2w + 2h, which represents the material needed for the tank's sides. We minimize P(w) = 2w + 2h subject to wh = 36, which allows us to write h = 36/w and minimize P(w) = 2w + 2(36/w) = 2w + 72/w. Choice D incorrectly maximizes area, but area is fixed by the constraint. The key insight is that when one geometric property is constrained, optimizing another property (like perimeter) determines the optimal shape.

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