Quadratic Models and Optimization (Max/Min) - College Algebra

1. A farmer wants to create a rectangular pen using 200 feet of fencing. One side of the pen will be against an existing barn wall, so fencing is only needed for three sides. If the side parallel to the barn has length $x$ feet, what value of $x$ maximizes the area of the pen?
   Answer: $x = 100$ feet Let the side parallel to the barn be $x$ feet and the perpendicular sides be $y$ feet each. The constraint is $x + 2y = 200$, so $y = \frac{200-x}{2}$. The area is $A = xy = x \cdot \frac{200-x}{2} = \frac{200x - x^2}{2} = -\frac{1}{2}x^2 + 100x$. The maximum occurs at $x = -\frac{100}{2(-1/2)} = 100$ feet. Choice A uses $x = 50$ (half the correct answer). Choice C uses 75 (incorrectly assumes equal sides). Choice D comes from incorrectly setting up the constraint as $2x + y = 200$.
2. A ball is thrown upward from the top of a 96-foot building with an initial velocity of 80 feet per second. The height $h$ (in feet) is given by $h(t) = -16t^2 + 80t + 96$. How long does it take for the ball to reach its maximum height, and when does it hit the ground?
   Answer: Maximum at $t = 2.5$ seconds; hits ground at $t = 6$ seconds When you encounter a projectile motion problem with a quadratic height function, you're dealing with a parabola that opens downward. The maximum height occurs at the vertex, and the ball hits the ground when the height equals zero. To find the maximum height time, use the vertex formula for a parabola $h(t) = at^2 + bt + c$. The time at the vertex is $t = -\frac{b}{2a}$. Here, $a = -16$ and $b = 80$, so $t = -\frac{80}{2(-16)} = -\frac{80}{-32} = 2.5$ seconds. To find when the ball hits the ground, set $h(t) = 0$ and solve: $-16t^2 + 80t + 96 = 0$. Dividing by $-16$ gives $t^2 - 5t - 6 = 0$. Factoring: $(t - 6)(t + 1) = 0$, so $t = 6$ or $t = -1$. Since time must be positive, the ball hits the ground at $t = 6$ seconds. Choice A incorrectly states the maximum occurs at $t = 5$ seconds, likely confusing the coefficient $b = 80$ with the vertex calculation. Choice B correctly finds the maximum time but incorrectly claims the ball hits ground at $t = 5$ seconds—this might come from misapplying the vertex time. Choice D has the right maximum time but uses $t = 4.8$ seconds for ground impact, possibly from a calculation error in the quadratic formula. Remember: for projectile problems, always use the vertex formula for maximum height and set the function equal to zero to find when the object hits the ground.
3. If one dimension of the rectangular flower bed is $x$ feet, what value of $x$ will maximize the area of the flower bed?
   Answer: $x = 15$ feet This is a classic optimization problem where you need to maximize area subject to a constraint. When you see "maximize area" with a fixed perimeter or border cost, think about setting up equations and using calculus or algebraic techniques to find the maximum. First, let's establish the constraint. The border costs $8 per linear foot with a $480 budget, so the perimeter is $480 ÷ 8 = 60$ feet. For a rectangle with dimensions $x$ and $y$, the perimeter formula gives us: $2x + 2y = 60$, which simplifies to $y = 30 - x$. The area function becomes $A = xy = x(30 - x) = 30x - x^2$. To maximize this quadratic function, you can complete the square or use calculus. Taking the derivative: $A' = 30 - 2x$. Setting this equal to zero gives $30 - 2x = 0$, so $x = 15$ feet. Since the second derivative is negative, this is indeed a maximum. Looking at the wrong answers: (A) $x = 7.5$ gives area $7.5 × 22.5 = 168.75$ square feet. (B) $x = 30$ would make the other dimension zero ($y = 0$), creating no garden at all. (D) $x = 20$ gives area $20 × 10 = 200$ square feet. The correct answer (C) $x = 15$ yields the maximum area of $15 × 15 = 225$ square feet. Study tip: For rectangular optimization problems with fixed perimeter, the maximum area always occurs when the rectangle is a square. Remember this shortcut to check your work!
4. A projectile is launched upward from a platform 64 feet high. Its height $h$ (in feet) after $t$ seconds is given by $h(t) = -16t^2 + 48t + 64$. At what time does the projectile reach its maximum height, and what is that maximum height?
   Answer: Maximum at $t = 1.5$ seconds; height = 100 feet For the quadratic $h(t) = -16t^2 + 48t + 64$, the maximum occurs at $t = -\frac{b}{2a} = -\frac{48}{2(-16)} = 1.5$ seconds. The maximum height is $h(1.5) = -16(1.5)^2 + 48(1.5) + 64 = -36 + 72 + 64 = 100$ feet. Choice B incorrectly uses $t = 3$ (double the correct time). Choice C has the correct time but miscalculates the height. Choice D uses $t = 2$ and gets the corresponding height at that incorrect time.
5. A company's profit $P$ (in dollars) from producing $x$ items is given by $P(x) = -3x^2 + 240x - 3000$. The company can only produce whole numbers of items. What production level yields the maximum profit, and what is that maximum profit?
   Answer: $x = 40$ items; profit = $1200 The vertex occurs at $x = -\frac{240}{2(-3)} = 40$ items. The maximum profit is $P(40) = -3(40)^2 + 240(40) - 3000 = -4800 + 9600 - 3000 = 1200$ dollars. Since 40 is a whole number, no rounding is needed. Choice A has the correct production level but miscalculates the profit. Choice C uses $x = 80$ (double the correct amount). Choice D attempts unnecessary rounding to $x = 39$ and computes $P(39)$.
6. The revenue $R$ (in thousands of dollars) for a company is modeled by $R(x) = -0.5x^2 + 8x + 12$, where $x$ is the number of units produced (in thousands). Due to production constraints, the company can produce at most 12,000 units. What is the maximum revenue the company can achieve?
   Answer: $44,000 The vertex occurs at $x = -\frac{8}{2(-0.5)} = 8$ thousand units. Since $8 < 12$, this is within the constraint. The maximum revenue is $R(8) = -0.5(8)^2 + 8(8) + 12 = -32 + 64 + 12 = 44$ thousand dollars = $44,000. Choice B evaluates $R$ at the constraint boundary $x = 12$. Choice C creates confusion about whether the constraint matters. Choice D results from arithmetic errors in computing $R(8)$.
7. A small business manufactures decorative boxes. The profit $P$ (in dollars) from selling $x$ boxes per week is modeled by $P(x) = -2x^2 + 160x - 1800$. If the business must sell at least 20 boxes per week to cover fixed costs, what is the maximum weekly profit the business can achieve?
   Answer: $1200 To find maximum profit, we need the vertex of the parabola $P(x) = -2x^2 + 160x - 1800$. The x-coordinate of the vertex is $x = -\frac{b}{2a} = -\frac{160}{2(-2)} = 40$. At $x = 40$: $P(40) = -2(40)^2 + 160(40) - 1800 = -3200 + 6400 - 1800 = 1200$. Since 40 ≥ 20, this satisfies the constraint. Choice A results from calculation errors in the vertex formula. Choice B comes from forgetting the constant term. Choice C is the result of sign errors when evaluating $P(40)$.

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