Question

Largest Rectangle A rectangle has its base on the \(x\) -axis and its upper two vertices on the parabola \(y=12-x^{2}\) . What is the largest area the rectangle can have, and what are its dimensions?

Short answer

The largest area the rectangle can have is 32 units. The dimensions of the rectangle are a height of 8 units and a width of 4 units.

Step-by-step solution

Define the function for the Area of the Rectangle

The upper vertices of the rectangle are on the parabola \(y = 12 - x^{2}\), thus the height of the rectangle will be the y-value of the parabola at any point x. Since the base of the rectangle is on the x-axis, it will be 2x wide (it extends x to the left and x to the right of the y-axis). So, the area of the rectangle \(A\) can be defined as the product of its height and its base which is \(A(x) = 2x(12-x^{2}) = 24x - 2x^{3}\).

Find the First Derivative and Set it to Zero

To find the maximum area of a rectangle, the derivative of the function \(A(x)\) needs to be computed and set to zero. The derivative of the function \(A(x) = 24x - 2x^{3}\) is \(A'(x) = 24 - 6x^{2}\). Setting \(A'(x) = 0\) results in the equation \(24 - 6x^{2} = 0\) which will give the values of x where the maximum or minimum can occur.

Solve for x-values

Solving the equation \(24 - 6x^{2} = 0\), one gets \(x^{2} = 4\), hence \(x = \pm 2\). Since the width of the rectangle (2x) must be positive, the only relevant solution for the x-value is \(x = 2\).

Find the Maximum Value of the Area

The maximum area can be found by substituting \(x = 2\) into the area function \(A(x)\). When \(x = 2\), \(A(2) = 2*2(12 - (2)^2) = 32\).

Find the Dimensions of Rectangle

The height of the rectangle at \(x = 2\) is the y-value of the parabola \(y = 12 - x^{2}\), thus \(y=12-2^{2}=8\), and the base of the rectangle, as previously noted, is \(2x=2*2=4\). Thus, the height is 8 units and the width is 4 units.

Key concepts

Derivative to Find Maximum

One of the most powerful applications of differentiation in calculus is finding the maxima or minima of a function. This process often involves taking the derivative of a function and setting it equal to zero to find critical points. In the context of optimizing the area of a geometric shape, such as a rectangle, this method is invaluable.

The first derivative of the area function, denoted as \( A'(x) \), represents the rate at which the area changes with respect to the width of the rectangle. By setting \( A'(x) = 0 \), we identify the points where the rate of change is zero – typically where the function switches from increasing to decreasing or vice versa. These points are possible candidates for maximum or minimum values. After finding these critical points, we can determine which represent the maximum area by evaluating the second derivative or by using a test, such as the First Derivative Test.

Area of a Rectangle

The area of a rectangle is a fundamental concept that is frequently used in optimization problems. It is simply the product of the rectangle's length and width. In formal terms, if a rectangle has a width \( w \) and a height \( h \), its area \( A \) is calculated as \( A = w \times h \).

In the case of our optimization problem, because the rectangle's vertices lie on a parabola and its base is on the x-axis, its height varies according to the y-value of the parabola at a given x-coordinate. Consequently, its area depends on a single variable, \( x \), after incorporating the parabolic relation. The simplicity of this calculation allows us to easily represent the area as a function of the variable and thus use calculus to optimize it.

Applications of Differentiation

Differentiation is a cornerstone of calculus with myriad applications in mathematics, physics, engineering, and economics. It allows us to analyze the rate at which things change and to optimize variables to achieve the best possible outcome in a given scenario.

Aside from finding maxima and minima, as in the example of maximizing the area of a rectangle, differentiation can help determine velocity and acceleration from a position function in physics, calculate the marginal cost or revenue in economics, and solve complex problems in engineering that involve rates of heat transfer, material stress, or fluid dynamics. The ability to translate real-world problems into functions and then use differentiation to analyze and optimize these functions is an essential skill in many fields.

Quadratic Functions

Quadratic functions are polynomial functions of the form \( f(x) = ax^2 + bx + c \), where \( a eq 0 \). They graph as parabolas, which can open upwards or downwards depending on the sign of the coefficient \( a \). In the case of our optimization problem, the parabola \( y = 12 - x^2 \) opens downwards, indicating that the maximum point of the function will give us the maximum height of the rectangle.

Quadratic functions are particularly important in optimization problems because they often represent relations in physical problems, such as the trajectory of projectiles or the distribution of stresses across surfaces. In our context, understanding the nature of the quadratic function allowed us to model the height of the rectangle on the curve and use differentiation to find the maximum area.