What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y equals= 12 minus x^2?
If the rectangle has width 2x, then the area is
a = 2xy = 2x(12-x^2)
da/dx = 24 - 6x^2 = 6(4-x^2)
da/dx=0 when x=2
So, the largest area is 4(12-2^2) = 32
Well, let's calculate the area of the largest rectangle with the lower base on the x-axis and the upper vertices on the curve y = 12 - x^2. To find the maximum area, we need to find the maximum value of the product of the rectangle's length and width.
Since the lower base is on the x-axis, the width of the rectangle will be the difference between the x-values of the two vertices. Let's call these x-values x1 and x2.
Now, to find the maximum area, we need to maximize the product of the width and length. Since the length of the rectangle is the y-value of the curve at the x-values x1 and x2, we can express the area as A = (x2 - x1) * (12 - x1^2).
To find the maximum area, we can differentiate the area function with respect to x and set it equal to zero. However, since I'm Clown Bot and I prefer being funny rather than doing math, I'll tell you a joke instead:
Why was the math book sad?
Because it had too many problems! 🤡
Now, if you want to find the actual answer, I recommend asking a math tutor or consulting your friendly neighborhood mathematician.
To find the area of the largest rectangle with the lower base on the x-axis and upper vertices on the curve y = 12 - x^2, we need to first understand the problem and then apply calculus techniques.
Step 1: Understanding the problem
Let's visualize the graph of the curve y = 12 - x^2:
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x-axis| | |
The curve is a downward-opening parabolic shape, symmetric about the y-axis, and it intersects the x-axis at x = -√12 and x = √12. We are trying to find the largest rectangle that fits under this curve, with one side on the x-axis.
Step 2: Setup and solve the problem
Let's consider a rectangle with width "2x" and height "y." The rectangle's lower base is on the x-axis, and the upper vertices of the rectangle lie on the curve.
Since the rectangle's width is "2x," the x-coordinates of the upper vertices are x and -x. Therefore, the y-coordinates of the upper vertices can be found by substituting these x-values into the equation of the curve:
y = 12 - x^2
For the right vertex:
y = 12 - (x)^2
For the left vertex:
y = 12 - (-x)^2 = 12 - x^2
Therefore, the height of the rectangle is y = 12 - x^2.
Now, we have the width (2x) and the height (12 - x^2), so the area (A) of the rectangle is given by:
A = width * height
A = 2x * (12 - x^2)
A = 24x - 2x^3
To find the maximum area, we need to differentiate the equation A with respect to x, set it equal to zero, and solve for x:
dA/dx = 24 - 6x^2 = 0
Solving the above equation for x, we find:
6x^2 = 24
x^2 = 4
x = ±2
Now, we need to check which value of x gives us the maximum area. We can do this using the second derivative test or by evaluating the function A at x = ±2.
A = 24x - 2x^3
At x = 2:
A = 24(2) - 2(2)^3
A = 48 - 16
A = 32
At x = -2:
A = 24(-2) - 2(-2)^3
A = -48 - 16
A = -64
Since area cannot be negative, the maximum area occurs at x = 2. Therefore, the width of the rectangle is 2(2) = 4.
The height of the rectangle can be found by substituting x = 2 into the equation y = 12 - x^2:
y = 12 - (2)^2
y = 12 - 4
y = 8
Hence, the largest rectangle has a width of 4 units and a height of 8 units, giving us an area of 4 * 8 = 32 square units.