solve for the max area of a rectangle inscribed in a circle with a radius of 8

Question

thanks !

Answer 1

The area of the inscribed rectangle is (2x)(2y) = 4xy. We wish to maximize this expression given the constraint x²+y² = 8² = 64.

One way to relate xy and x²+y² is by using the perfect square formula (x-y)² = x²+y²-2xy.

Since a perfect square must be non-negative, we have 0 ≤ (x-y)² = x²+y²-2xy, which gives us 2xy ≤ x²+y² = 64.

Therefore, the area of the rectangle is bounded by 4xy ≤ 128.

To prove that this is indeed the maximum area, we have to show that this number can be achieved. We go back to the perfect square, which must be 0 for equality to occur. That is when x-y = 0, or equivalently x = y. Plug this into the constraint equation and we have 2x² = 64, i.e. x = 4√2.

Answer 2

If you sketch out a circle at the origin of radius "r", with an inscribed rectangle, you get something like this:

The graph of the circle is x² + y² = r² or in this case, x² + y² = 8² or x² + y² = 64 which means y = √(64 - x²)

The y-value of the corner point (upper right) is "y", which makes the height of the rectangle 2y. The x-value of the upper right corner is "x" which makes the width of the rectangle 2x.

Then the area of the rectangle is given by:

A = (2x)(2y) = 4xy

Substituting in y = √(64 - x²) gives us:

A(x) = 4x√(64 - x²) or A(x) = 4x(64 - x²)^1/2

To find the maximum, take the derivative and set it equal to zero:

To take the derivative, you can use the product rule (u•v)' = u'v + uv'

u = 4x so u' = 4

v = (64 - x²)^1/2 so v' = 1/2((64 - x²)^-1/2(-2x) using the chain rule or, simplifying: v' = -x/(64 - x²)^1/2

Putting these together we get:

A'(x) = 4(64 - x²)^1/2 + (4x)(-x/(64 - x²)^1/2)

A'(x) = 4(64 - x²)^1/2 - 4x²/(64 - x²)^1/2

4(64 - x²)^1/2 - 4x²/(64 - x²)^1/2 = 0

4(64 - x²)^1/2 = 4x²/(64 - x²)^1/2

(64 - x²)^1/2 = x²/(64 - x²)^1/2

Cross multiplying:

(64 - x²)^1/2(64 - x²)^1/2 = x²(1)

64 - x² = x²

64 = 2x²

32 = x²

x = ± √32

x = ± 4√2 (you can use the first derivative test to verify that this is a maximum)

We can disregard the negative answer since we are thinking of the upper right corner

So x = 4√2

Since y = √(64 - x²) then y = √(64 - (4√2)²) = √(64 - 32) = √32 = 4√2

Then, calculating the area: A = 4xy = 4(4√2)(4√2) = 128 square units