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A rectangle is bounded by the x-axis and the semicircle y=�ã(25-x^2).

Question is, what length and width should the rectangle have so that its area is a maximum, and what is the maxuimum area?


Area= length*width
= 2x*y= 2x*sqrt(25-x^2)

Now, take that, differentiate it, set to zero, and solve for x,y. Length = 2x, width (or height) is y.
I will be happy to critique your work or thinking.

User Icon for Bot Bot answered
1 year ago

Differentiating with respect to x:

2*sqrt(25-x^2) - 4x*(1/2)*(25-x^2)^(-1/2) = 0

2*sqrt(25-x^2) = 4x*(1/2)*(25-x^2)^(-1/2)

4x^2 = 25-x^2

5x^2 = 25

x = sqrt(25/5)

y = sqrt(25-x^2)

= sqrt(25-(25/5))

= sqrt(20/5)

Therefore, the length and width of the rectangle should be 2*sqrt(25/5) and sqrt(20/5) respectively, and the maximum area is (2*sqrt(25/5))*(sqrt(20/5)) = 4*sqrt(500/25) = 4*sqrt(20) units^2.

User Icon for Explain Bot Explain Bot answered
1 year ago

To find the dimensions that maximize the area of the rectangle, we need to maximize the function representing the area. In this case, the area function is given by:

Area = 2x * √(25 - x^2)

To find the maximum area, we can differentiate the area function with respect to x and set it equal to zero.

d(Area)/dx = 2 * √(25 - x^2) + 2x * (1/2) * (25 - x^2)^(-1/2) (-2x)

Simplifying this expression, we get:

2 * √(25 - x^2) - (2x^2) / √(25 - x^2) = 0

Multiplying through by √(25 - x^2), we get:

2(25 - x^2) - 2x^2 = 0

Expanding and rearranging, we have:

50 - 2x^2 - 2x^2 = 0

4x^2 = 50

x^2 = 50/4

x^2 = 12.5

Taking the positive square root (since dimensions cannot be negative), we have:

x = √12.5

Now, we can substitute this value of x back into the area function to find the corresponding value of y:

Area = 2x * √(25 - x^2)

Substituting x = √12.5, we have:

Area = 2 * √12.5 * √(25 - 12.5)

Area = 2 * √(12.5) * √(12.5)

Area = 2 * 3.5355 * 3.5355

Area ≈ 24.999

Therefore, the maximum area of the rectangle is approximately 24.999 square units. The length and width of the rectangle are approximately equal to 2√12.5 units.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To find the dimensions of the rectangle that would maximize its area, we need to differentiate the equation for the area with respect to x and set it equal to zero.

Let's start by differentiating the expression for the area:

A = 2x * sqrt(25 - x^2)

To simplify the differentiation process, let's rewrite the expression as:

A = 2x * (25 - x^2)^(1/2)

Now, using the product rule of differentiation, we can find the derivative of A with respect to x:

dA/dx = 2 * sqrt(25 - x^2) - 2x * (1/2) * (25 - x^2)^(-1/2) * (-2x)

Simplifying this expression further:

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

Setting this derivative equal to zero to find the critical points:

2 * sqrt(25 - x^2) + x^2 / sqrt(25 - x^2) = 0

Multiplying both sides of the equation by sqrt(25 - x^2):

2(25 - x^2) + x^2 = 0

Expanding and simplifying:

50 - 2x^2 + x^2 = 0

2x^2 = 50

x^2 = 25

Taking the square root of both sides and considering only the positive square root:

x = 5

Now that we have found the critical point x = 5, we can substitute this value back into the equation for the area to find the corresponding value of y:

A = 2x * sqrt(25 - x^2)

A = 2 * 5 * sqrt(25 - 5^2)

A = 10 * sqrt(25 - 25)

A = 0

The area of the rectangle at the critical point x = 5 is zero, which does not make sense in this context. Therefore, we need to consider the endpoints of the interval as well.

Since the rectangle is bounded by the x-axis and the semicircle, the possible values for x are in the interval [-5, 5].

First, let's consider the value of x = -5:

A = 2 * (-5) * sqrt(25 - (-5)^2)

A = -10 * sqrt(25 - 25)

A = -10 * sqrt(0)

A = 0

Next, let's consider the value of x = 5:

A = 2 * 5 * sqrt(25 - 5^2)

A = 10 * sqrt(25 - 25)

A = 0

So, at both endpoints of the interval, the area of the rectangle is zero.

Therefore, there is no maximum area for the rectangle within the given domain.