Problem 74
Question
A rectangular garden plot is to be enclosed with a fence on three of its sides and a brick wall on the fourth side. If 100 feet of fencing material is available, what dimensions will yield the maximum area? The height of a ball that is thrown directly upward from a point 200 feet above the ground with an initial velocity of 40 feet per second is given by \(h(t)=-16 t^{2}+40 t+200,\) where \(t\) is the amount of time elapsed since the ball was thrown. Here, \(t\) is in seconds and \(h(t)\) is in feet. (a) Sketch a graph of \(h\) (b) When will the ball reach its maximum height, and what is the maximum height?
Step-by-Step Solution
Verified Answer
The dimensions that will yield the maximum area for the garden are 25 feet by 50 feet with a maximum area of 625 square feet. The ball will reach its maximum height of 225 feet at 1.25 seconds.
1Step 1: Setup Function for Area
Let \(x\) denote the width of the garden and \(y\) be the length. Formulate the function \(A(x) = x(100 - 2x)\) where \(A(x)\) represents the area of the garden.
2Step 2: Find Maximum Area
To find the maximum area, differentiate \(A(x)\) with respect to \(x\) and set it to 0 to get critical points. \(A'(x) = 100 - 4x\), setting \(A'(x) = 0\) yields \(x = 25\). Substituting \(x = 25\) into \(A(x)\) gives maximum area \(A(max) = 625\) square feet.
3Step 3: Setup for Ball's Height Function
The ball's height function is already given by \(h(t)=-16 t^{2}+40 t+200\).
4Step 4: Graph the Function
Sketch the function \(h(t)\) against time \(t\). This will generate a downward opening parabola as the coefficient of \(t^{2}\) is negative.
5Step 5: Maximum Ball's Height
The maximum height of the ball is achieved at the vertex of the parabola. The time at which it achieves this can be calculated by \(t = -b/2a = -40/-32 = 1.25\) seconds. Substituting \(t = 1.25\) into \(h(t)\) yields maximum height \(h(max) = 225\) feet.
Key Concepts
Maximum Area OptimizationQuadratic FunctionsApplication of Derivatives
Maximum Area Optimization
Imagine you're helping a friend plan a garden. They've got a fence along one side and need to use the remaining material to enclose the largest possible space. It's like a puzzle, where every piece is a choice that controls the shape and size of the garden. The problem here is strikingly similar; it’s all about making the right choices to optimize for the maximum area.
Let's start with the principle that to maximize the area of a rectangle with a given perimeter is to aim for a square shape. But here, only three sides will have fencing. So, the strategy shifts slightly—two sides will be equal (those opposite each other), and the third will adjust so that the total length doesn’t exceed 100 feet. This is where calculus comes into play. By setting up the area function given the perimeter constraint and using calculus to find when our area function reaches its peak, we zero in on the optimal dimensions.
For those not mathematically inclined, the idea is simple: by playing around with the dimensions within our limit (100 feet of fencing), we can figure out the best width and length that give us the biggest garden. This method is handy and goes beyond just garden layouts—it's a way of making the best out of any limited resource, whether it’s time, money, or, in this case, fencing.
Let's start with the principle that to maximize the area of a rectangle with a given perimeter is to aim for a square shape. But here, only three sides will have fencing. So, the strategy shifts slightly—two sides will be equal (those opposite each other), and the third will adjust so that the total length doesn’t exceed 100 feet. This is where calculus comes into play. By setting up the area function given the perimeter constraint and using calculus to find when our area function reaches its peak, we zero in on the optimal dimensions.
For those not mathematically inclined, the idea is simple: by playing around with the dimensions within our limit (100 feet of fencing), we can figure out the best width and length that give us the biggest garden. This method is handy and goes beyond just garden layouts—it's a way of making the best out of any limited resource, whether it’s time, money, or, in this case, fencing.
Quadratic Functions
Move over, linear equations, quadratic functions are here to show how things can get curvy. These functions pack a punch with their distinctive 'U' shape parabolas, either smiling up at the sky or frowning towards the ground.
Speaking in algebra terms, a quadratic function takes on the general form of \(y=ax^2+bx+c\), with the variables 'a', 'b', and 'c' shaping the parabola's width, direction, and position. When the leading coefficient 'a' is positive, the parabola opens upwards like a welcoming bowl. If 'a' is negative, the parabola is like an umbrella shielding us from rain, pointing downwards.
To connect this with our garden problem, there isn’t an obvious parabola to speak of. But these functions are everywhere, including in trajectories, like the path of a tossed ball—our second problem within this exercise. The height of the ball, tossed with a certain force, creates a perfect scenario to observe a quadratic function in real-life action. The downward sweep of the graph as time ticks shows the ball's ascent and descent, deftly illustrated by a quadratic function's predictable curve.
Speaking in algebra terms, a quadratic function takes on the general form of \(y=ax^2+bx+c\), with the variables 'a', 'b', and 'c' shaping the parabola's width, direction, and position. When the leading coefficient 'a' is positive, the parabola opens upwards like a welcoming bowl. If 'a' is negative, the parabola is like an umbrella shielding us from rain, pointing downwards.
To connect this with our garden problem, there isn’t an obvious parabola to speak of. But these functions are everywhere, including in trajectories, like the path of a tossed ball—our second problem within this exercise. The height of the ball, tossed with a certain force, creates a perfect scenario to observe a quadratic function in real-life action. The downward sweep of the graph as time ticks shows the ball's ascent and descent, deftly illustrated by a quadratic function's predictable curve.
Application of Derivatives
Derivatives aren't just for financial wizards; they are the mathematicians' secret weapon for understanding how things change. Simply put, if you have a graph of something moving or growing, the derivative is like a speedometer that tells you how fast it's going. It's a snapshot of the rate of change at any given moment.
When applied to optimization problems, derivatives provide a way to pinpoint where a function — like our area function for the garden plot — hits its maximum or minimum. It's like climbing a hill: at the top, your elevation doesn't change instantaneously; you have a moment where you're neither going up nor down. That's the 'critical point' and derivatives help us find it by setting them equal to zero.
In the case of our thrown ball, the derivative of the height function indicates the velocity of the ball at any second. When it reaches its peak height, the rate of ascent drops to zero, signifying a 'critical point'—that fleeting instant the ball hangs in the air before gravity pulls it back down. It’s through derivatives that we capture that moment and mathematically show the exact height and time of the ball's maximum altitude. These concepts demonstrate the beauty of calculus in motion, giving us not just the 'how fast?' but the 'when?' and 'how high?'
When applied to optimization problems, derivatives provide a way to pinpoint where a function — like our area function for the garden plot — hits its maximum or minimum. It's like climbing a hill: at the top, your elevation doesn't change instantaneously; you have a moment where you're neither going up nor down. That's the 'critical point' and derivatives help us find it by setting them equal to zero.
In the case of our thrown ball, the derivative of the height function indicates the velocity of the ball at any second. When it reaches its peak height, the rate of ascent drops to zero, signifying a 'critical point'—that fleeting instant the ball hangs in the air before gravity pulls it back down. It’s through derivatives that we capture that moment and mathematically show the exact height and time of the ball's maximum altitude. These concepts demonstrate the beauty of calculus in motion, giving us not just the 'how fast?' but the 'when?' and 'how high?'
Other exercises in this chapter
Problem 73
Find all solutions of the quadratic equation. Relate the solutions of the equation to the zeros of an appropriate quadratic function. $$\frac{2}{3} x^{2}+x=-1$$
View solution Problem 74
Sketch a graph of the quadratic function, indicating the vertex, the axis of symmetry, and any \(x\)-intercepts. $$g(t)=3 t^{2}-6 t-\frac{3}{4}$$
View solution Problem 74
In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=3 x+1 ; g(x)=
View solution Problem 74
Use a graphing utility to solve the problem. If \(f(x)=\sqrt{x},\) graph \(3 f(x)\) and \(f(3 x)\) in the same viewing window. Are the graphs the same? Explain.
View solution