Find the largest area of the largest rectangle that can be cut from a circular quadrant of diameter 8 feet.

Question

Please show your solution on how you arrived with the answer. Thank you.

Answer 1

That question seems to mean you have 1/4 of a circle, where the circle has diameter 8 feet

The quadrant has a width and height of 4 feet, with an arc of length pi/2 x 4 = 2pi

The equation for a circle is x^2 + y^2 = r^2 = 4^2 = 16

For just the 1st quadrant the equation of the 1/4 circle can be

solved for y = +square root of (16-x^2) where the domain of x and y is from 0 to 4

For the area function you want xy=Area = xsqr(16-x^2)

Take the derivative of xy and set it equal to zero

(xy)'= x(1/2)(16-x^2)^1/2 times -2x + y

=-x^2/sqr(16-x^2) + sqr(16-x^2) = 0

multiply by sqr(16-x^2)

= -x^2 + 16 -x^2 = 0

solve for x

-2x^2 +16 = 0

2x^2 = 16

x^2 = 8

x= sqr8

y=sqr(16-x^2)

=sqr(16-8)

=sqr(8)

the area with maximum area is sqr8 by sqr8 = 8 square feet

It's a rectangle that is a square

with all 4 sides = sqr8 = about 2.828 feet

= approximately 2 feet and 9.94 inches

The circle quadrant intersects or just touches the rectangle at the point (x,y) = (sqr8, sqr8)

Answer 2

To start, draw a rectangle inscribed in a circle and connect two of the vertices through the center of the circle. you'll notice that the diagonal going through the rectangle is also the diameter of the circle. This will help us relate the side length dimensions of the rectangle to the diameter of the circle. Label one side of the rectangle x. Now we have x as one side length and the diameter (d) as the diagonal and we can use pythagorean theorem to solve for the other side of the rectangle in terms of x and d. Now you have the sides of the rectangle as:

length = x

width = √(d² - x²) = √(8² - x²) = √(64 - x²)

Write an expression for the area:

A = length * width = x * √(64 - x²) = x * (64 - x²)^1/2

Find the derivative of the area equation using the product rule and chain rule:

A' = [1 * (64 - x²)^1/2] + [x * (1/2)(64 - x²)^-1/2 (-2x)] = (64 - x²)^1/2 - x²(64 - x²)^-1/2

For optimization problems, set A' equal to zero and solve for x:

A' = (64 - x²)^1/2 - x²(64 - x²)^-1/2 = 0

(64 - x²)^1/2 = x²(64 - x²)^-1/2

[(64 - x²)^1/2] / [(64 - x²)^-1/2] = x²

64 - x² = x²

64 = 2x²

32 = x²

x = ±√32

Obviously a length can't be negative, so we reject -√32 and x = +√32

This is the largest value of x. To find the largest area, plug x = +√32 into the original equation for A:

A = x * (64 - x²)^1/2 = (√32) * (64 - (√32)²)^1/2 = √32 * (64 - 32)^1/2 = √32 * (32)^1/2 = √32 * √32 = 32 ft²