A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y= 12 - x^2. What are the dimensions of such a rectangle with the greatest possible area?
The vertices of the rectangle are points:
(-x,0),(x,0),(x,12-x^2),(-x,12-x^2).
The area A(x)=2x(12-x^2)
A(x)=24x-2x^3
A'(x)=24-6x^2=0 => x=+-2
max A(x)=A(2)=48-16=32
To find the dimensions of the rectangle with the greatest possible area, we can start by visualizing the problem.
First, let's assume that the base of the rectangle is located on the x-axis. Since the x-axis corresponds to the width of the rectangle, we can call this dimension "w".
Now, let's consider the height of the rectangle. The upper corners of the rectangle are on the parabola y = 12 - x^2, so the height is given by the difference between the y-values of these two points. We can call this dimension "h".
To find the dimensions of the rectangle with the greatest possible area, we need to maximize the area A = w * h.
Let's break the problem down into steps:
1. Express the height h in terms of x:
Since the upper corners of the rectangle are on the parabola y = 12 - x^2, the y-values of these points are given by y = 12 - x^2. So, the height h = 12 - x^2.
2. Express the area A in terms of x:
The area of the rectangle is calculated as A = w * h. Substituting the expressions for height and width, we get A = w * (12 - x^2).
3. Express the area A in terms of a single variable:
Since the question asks for the dimensions with the greatest possible area, we want to express the area A in terms of a single variable. In this case, we can express it in terms of the variable x.
4. Maximize the area A:
To maximize the area, we need to find the value of x that maximizes A. Our expression for the area A is A = w * (12 - x^2), but we know that the width w is located on the x-axis, so it must always be positive. Therefore, we need to find the maximum value of (12 - x^2) while x varies.
The parabola y = 12 - x^2 is a downward-opening parabola, and its maximum point occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula x = -b / 2a, where the quadratic equation is in the form ax^2 + bx + c = 0. In our case, a = -1, b = 0, and c = 12, so the x-coordinate of the vertex is x = -0 / (2 * -1) = 0.
Therefore, the maximum value of (12 - x^2) occurs at x = 0, and the maximum area A is given by A = w * (12 - 0^2) = 12w.
5. Find the dimensions of the rectangle:
Now that we know the maximum area A is given by 12w, we need to find the value of w (width) that maximizes the area. Since the width w is located on the x-axis and it must be positive, the dimensions of the rectangle with the greatest possible area are w = 0 and h = 12 - 0^2 = 12.
So, the dimensions of the rectangle with the greatest possible area are 0 for the width (w) and 12 for the height (h).