Find the Area of a Rectangle Bounded by a Semicircle | Geometry Problem Solution

[slayer16]

Homework Statement

A rectangle is bounded by the x-axis and the semicircle y=√(36-x^2) (see figure). Write the area A of the rectangle as a function of x.

I will try to explain the figure as best as possible. The figure is basically a semicircle on the Cartesian plane with a domain of [-6,6]. There is a point on the semicircle of (X,Y) where y is the height of the rectangle and x is the maximum width of the rectangle. The value of x is around five on the graph, while the value for y is around 3.5 on the graph (note: they are just estimates to help visualize the figure better).

Homework Equations

A=(w)(h)

The Attempt at a Solution

I substitute the (x,y) for (w,h). I then realized that the height stays constant, therefore I can write the height as √(36-w^2), where w is a constant. Now, I do not know how to write the width because the x has a limited domain.

 

[jegues]

Couldn't you use a definite integral to evaluate this?

 

[slayer16]

Couldn't you use a definite integral to evaluate this?

Not really. The question is found in a precalculus textbook.

 

[LCKurtz]

Homework Statement

A rectangle is bounded by the x-axis and the semicircle y=√(36-x^2) (see figure). Write the area A of the rectangle as a function of x.

I will try to explain the figure as best as possible. The figure is basically a semicircle on the Cartesian plane with a domain of [-6,6]. There is a point on the semicircle of (X,Y) where y is the height of the rectangle and x is the maximum width of the rectangle. The value of x is around five on the graph, while the value for y is around 3.5 on the graph (note: they are just estimates to help visualize the figure better).

Homework Equations

A=(w)(h)

The Attempt at a Solution

I substitute the (x,y) for (w,h). I then realized that the height stays constant, therefore I can write the height as √(36-w^2), where w is a constant. Now, I do not know how to write the width because the x has a limited domain.

The rectangle with a corner at (x,y) on your circle in the first quadrant has a base of length 2x and its height is y = sqrt(36-x²). Multiply those together and you will have the area as a function of x. Apparently you are not asking for the dimensions of such a rectangle with largest area, just the expression for A as a function of x.

 

[slayer16]

The rectangle with a corner at (x,y) on your circle in the first quadrant has a base of length 2x and its height is y = sqrt(36-x²). Multiply those together and you will have the area as a function of x. Apparently you are not asking for the dimensions of such a rectangle with largest area, just the expression for A as a function of x.

Thanks LCKurtz