A rectangle is bounded by the x axis and the semicircle = square root 25-x^2. What length and width should the rectangle have so that its area is a maximum?

When I worked the problem out(which is a bit detailed). I started with y = square root 25-x^2 then A = 2x square root 25-x^2. For my answer, I got y = square root 25-25/2 = square root 25/2= 5 square root 2/2. Then A= 5 square root 2 times 5 square root 2/2 =25. But I don't know if my answer 25 is correct.

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Okay thanks :), that did help

To find the length and width that maximize the area of the rectangle, you need to maximize the area function A = 2x√(25 - x^2), where x represents the length of the rectangle.

To find the maximum value of this area function, we can take the derivative of A with respect to x and set it equal to zero.

dA/dx = 2(√(25 - x^2)) + 2x(-x/√(25 - x^2))
dA/dx = 2(√(25 - x^2)) - 2x^2/√(25 - x^2)

Setting this derivative equal to zero, we get:

2(√(25 - x^2)) - 2x^2/√(25 - x^2) = 0

Simplifying this equation and cross-multiplying, we get:

2(√(25 - x^2))^2 = 2x^2

4(25 - x^2) = 2x^2

100 - 4x^2 = 2x^2

6x^2 = 100

x^2 = 100/6 = 50/3

Taking the square root of both sides, we get:

x = √(50/3) = (5√2)/√3 = (5√2√3)/(√3√3) = (5√6)/3

Now that we have the value of x, we can substitute it back into the area function to find the maximum area:

A = 2x√(25 - x^2)
A = 2[(5√6)/3]√(25 - (5√6/3)^2)
A = (10√6)/3 * √(25 - 25/3)
A = (10√6)/3 * √(75/3 - 25/3)
A = (10√6)/3 * √(50/3)
A = (10√6)/3 * √(50)/(√3)
A = (10√6 * √50)/(3√3)
A = (10√6 * 5√2)/(3√3)
A = (50√12)/(3√3)
A = (50 * 2√3)/(3√3)
A = (100√3)/(3√3)
A = 100/3

So, the maximum area of the rectangle is 100/3, or approximately 33.33. Therefore, your answer of 25 is not correct. The correct maximum area is 100/3.

To find the length and width of the rectangle that maximize its area within the given constraints, let's break down the problem into steps:

Step 1: Define the dimensions of the rectangle.
Let the length of the rectangle be 2x (corresponding to the x-axis) and the width be y (which corresponds to the height of the semicircle).

Step 2: Determine the area of the rectangle.
The area (A) of a rectangle is given by multiplying its length by its width, so A = 2xy.

Step 3: Express the width (y) in terms of x.
We are given that the rectangle is bound by the x-axis and the equation of the semicircle: y = √(25 - x^2).

Step 4: Substitute the expression for y into the equation for area (A).
By substituting y = √(25 - x^2) into A = 2xy, we get:
A = 2x * √(25 - x^2).

Step 5: Maximize the area.
To find the maximum area, we differentiate A with respect to x, set the derivative equal to zero, and solve for x:

dA/dx = 0
2√(25 - x^2) - 2x * (1/2)(25 - x^2)^(-1/2) * (-2x) = 0
Simplifying this equation gives:
√(25 - x^2) + x^2/(√(25 - x^2)) = 0

Squaring both sides of the equation:
25 - x^2 + x^2 = 0
25 = x^2

Taking the positive square root:
x = 5

Step 6: Find the corresponding width (y).
Substituting x = 5 into the equation for y gives:
y = √(25 - 5^2) = √(25 - 25) = √0 = 0

Step 7: Calculate the area.
Substituting x = 5 and y = 0 into the equation for the area (A = 2xy):
A = 2 * 5 * 0 = 0

Therefore, the length and width of the rectangle that maximize its area within the given constraints are 2x = 2 * 5 = 10 and y = 0, respectively. However, since the width is 0, the area of the rectangle is zero as well.

In conclusion, there is no rectangle that satisfies the given constraints while maximizing the area.