Applications of derivatives

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.
Support your answer graphically.

Well, the parabola is symettric about x=0, so just solve the half of rectangle, with x=0 being the one side, and x the other.

Area= xy= x(12-x^2)= 12x-x^3

dA/dx= 12 -3x^2 or setting to zero, x=2

Now applying symettry, the rectangle is from x=-2 to x=2

Area= INt ydx from x=-2 to 2
INT (12-x^2) dx= 12x-1/3 x^3 limits..
Area= 12*2 - 1/3 (-8) + 12*2 +1/3(8)

I don't know what support graphically means, unless it means draw a graph.

check my work.

x = 4 and y = 8

the area = 32

To find the largest area of the rectangle and its dimensions, we need to maximize the area function, which in this case is given by A = x(12 - x^2), where x represents the length of the rectangle on the x-axis.

First, we find the derivative of the area function with respect to x, dA/dx. Taking the derivative, we have dA/dx = 12 - 3x^2.

To find the critical point(s) where the derivative equals zero, we set dA/dx = 0 and solve for x:

12 - 3x^2 = 0
3x^2 = 12
x^2 = 4
x = ±2

Since we are considering the positive range for the x-axis, the critical point is x = 2.

Now, we need to check if this critical point is a maximum or minimum by examining the second derivative, d^2A/dx^2. Taking the second derivative, we have d^2A/dx^2 = -6x.

Substituting x = 2 into the second derivative, we have d^2A/dx^2 = -6(2) = -12.

Since the second derivative is negative, this means the critical point x = 2 corresponds to a maximum.

Now, we need to determine the dimensions of the rectangle. Since we have only found the value for x, we can plug this value back into the area function to find the maximum area:

A = x(12 - x^2)
A = 2(12 - 2^2)
A = 2(12 - 4)
A = 2(8)
A = 16

Therefore, the largest area the rectangle can have is 16 square units. To determine the dimensions, we know that the base of the rectangle lies on the x-axis, so the length is 2 units. The height can be found by substituting x = 2 into the equation of the parabola: y = 12 - x^2. Therefore, the height is y = 12 - 2^2 = 8 units.

Graphically, you can plot the parabola y = 12 - x^2 and identify the points where the rectangle lies. The x-coordinate of the critical point, 2, will correspond to the bottom edge of the rectangle on the x-axis. The upper two vertices will have coordinates (2, 8) and (-2, 8).