A rectangle is bounded by the x-axis and the semicircle y = sqrt(36-x^2). What length and width should the rectangle have so that its area is a maximum?

I understand that 2xy = A and that 4x + 2y = P, but I'm not sure how to solve for a variable to plug back into the equation. I keep getting 6, but then the other variable would be 0.

Well, let's see if I can add some humor to help you solve this problem!

To find the dimensions of the rectangle, we need to maximize its area. Now, imagine the rectangle throwing a party and inviting all its shape friends. The rectangle wants to be the star of the show and have the largest area possible. So, it needs to find the perfect length and width to impress everyone!

Since the rectangle is bounded by the x-axis and the semicircle, we know that the x-coordinate on the graph will represent the width of the rectangle. Let's call it "w" for width (because "x" was busy being a variable for the x-coordinate).

Now, the y-coordinate on the graph represents the height of the rectangle, which we'll call "h" for humor (because, well, why not?).

To maximize the area of the rectangle, we need to set up an equation. Now, the area of a rectangle is just the product of its length and width, which in our case would be "wh". So, we want to maximize the function "A = wh".

But wait, we haven't used the equations you mentioned yet! Remember "2xy = A" and "4x + 2y = P" (where "P" is the perimeter of the rectangle)? Well, it's time for them to join the party!

Since we know the area is "wh", we can rewrite it as "2xy". So, "2xy = A". And because we want to maximize "A", we can rewrite the equation as "2xy = wh".

Now, let's get the other equation "4x + 2y = P" involved. But here's the thing, we want to express one variable in terms of the other to plug it back into our area equation. Let's pick "x" for convenience!

Solving the equation for "x" gives us "x = (P - 2y)/4". Now, let's substitute this into our area equation.

Replacing "x" with "(P - 2y)/4" gives us: "2(P - 2y)/4 * y = wh". Simplifying further, we have "(P - 2y) * y = 2wh".

Now it's time to unleash the humor side of the equation! Let's expand it: "Py - 2y^2 = 2wh". Rearranging terms, we have "-2y^2 + Py = 2wh".

Now, we have a quadratic equation in terms of "y"! It's time to make some laughs and solve it. Just keep in mind that we're looking for the maximum area, so we'll be using the vertex formula.

The formula for the x-coordinate of the vertex is given by "x = -b/(2a)". Applying this formula, we get "y = -P/(-4) = P/4". But hang on, we want the area to be a maximum, not minimum! So, let's give the equation a twist and flip it upside down.

Flipping the equation gives us "2y^2 - Py = -2wh". Rearranging again, we have "2y^2 - Py + 2wh = 0".

Using the vertex formula for the new equation, we find "y = -(-P)/(-4) = P/4". It might sound a bit negative, but this is good news because it represents the maximum area!

Finally, substitute "y" back into our equation "4x + 2y = P" to find "x": "4x + 2(P/4) = P". Simplifying gives us "4x + P/2 = P".

Rearranging terms, we have "4x = P - P/2", which simplifies to "4x = P/2". Solving for "x" gives us "x = P/8".

So, the width of the rectangle is "w = x = P/8", and the height is "h = y = P/4".

I hope all these shape parties and humor have helped you find the dimensions of the rectangle that maximize its area. Remember, when in doubt, bring out the jokes!

To find the dimensions of the rectangle that will maximize its area, we can use calculus.

Let's start by drawing the situation described. The rectangle is bounded by the x-axis and the semicircle y = sqrt(36-x^2).

Since the rectangle is bounded by the x-axis, the length of the rectangle will be 2x, and since the rectangle is bounded by the semicircle, the width will be y.

To find the area of the rectangle, we multiply the length and the width: A = 2x * y. We want to find the values of x and y that maximize the area A.

We also know the equation of the semicircle: y = sqrt(36 - x^2).

Now we can substitute the value of y into the equation for the area:
A = 2x * sqrt(36 - x^2).

To find the maximum area, we need to take the derivative of A with respect to x, set it equal to zero, and solve for x.

dA/dx = 2sqrt(36 -x^2) + 2x * (1/2) * (36 - x^2)^(-1/2) * (-2x)
= 2sqrt(36 - x^2) - 2x^2/sqrt(36 - x^2)

Setting dA/dx equal to zero, we get:
2sqrt(36 - x^2) - 2x^2/sqrt(36 - x^2) = 0.

Now we can solve this equation for x. Multiply both sides by sqrt(36 - x^2) to get rid of the denominators:
2(36 - x^2) - 2x^2 = 0
72 - 2x^2 - 2x^2 = 0
4x^2 = 72
x^2 = 18
x = sqrt(18) = 3sqrt(2).

Now we have the value of x. To find the value of y, we can substitute the value of x back into the equation of the circle:
y = sqrt(36 - (3sqrt(2))^2) = sqrt(36 - 18) = sqrt(18) = 3.

So the dimensions of the rectangle that will maximize its area are length = 2x = 2 * 3sqrt(2) = 6sqrt(2), and width = y = 3.

Therefore, the rectangle should have a length of 6sqrt(2) units and a width of 3 units to maximize its area.

To find the dimensions of the rectangle that will maximize its area, we need to use the given equations and optimize the area function.

Let's start by understanding the problem. We have a rectangle bounded by the x-axis and the semicircle y = sqrt(36 - x^2). This means that the base of the rectangle will lie on the x-axis, and the top side will be part of the semicircle.

We can represent the dimensions of the rectangle using variables. Let's call the length of the rectangle x, and the width y.

Now, we know that the area of a rectangle is given by A = Length x Width. In this case, A = xy.

We also have another equation that represents the perimeter of the rectangle. It says that 4x + 2y = P, where P is the perimeter of the rectangle.

To proceed, we need to eliminate one variable from the equations. Let's solve the second equation for x:

4x + 2y = P
4x = P - 2y
x = (P - 2y)/4

Now, substitute this expression for x into the area equation:

A = xy
A = (P - 2y)y/4
A = Py/4 - y^2/2

We now have the area A expressed solely in terms of y. To maximize the area, we will take the derivative of A with respect to y and set it equal to zero:

dA/dy = P/4 - 2y = 0

Solving this equation gives us:

P/4 - 2y = 0
2y = P/4
y = P/8

Now that we have found the value of y that maximizes the area, we can substitute it back into the equation for x:

x = (P - 2y)/4
x = (P - 2(P/8))/4
x = (P - P/4)/4
x = (3P/4)/4
x = 3P/16

Therefore, the length and width of the rectangle that maximize its area are (3P/16) and (P/8), respectively.

Note: It seems like you were trying to solve the problem by setting 2xy = A and 4x + 2y = P. While these equations are correct, the approach you described can lead to an incorrect result. Instead, it's important to express one of the variables in terms of the other and then substitute it back into the equation to optimize the area.

we don't care about P.

y = √(36-x^2)

a = 2xy = 2x√(36-x^2)
da/dx = 4(18-x^2)/√(36-x^2)

so, da/dx=0 when 18-x^2=0, or x = √18

now you can find y.