### The Archrival

Problem Translation:

A rectangle is inscribed in a circle with a radius of 4.

What is the major equation required for this problem?

Find the critical values of x

What is the largest area the rectangle can have and what are the dimensions?

ANSWERS:

The major equation required for this problem is

The area of the rectangle A = length * width

Since a circle is not a function, we use a semi-circle so that our calculations will be satisfied and more valid.

The equation of this circle is x2 + y2 = 4

So, the coordinates of the rectangle with respect to x are:

Length = 2x width = √ 16 – x2 formula – A(x) = x √ 16 – x2

Thus, the domain for this function would be [ 0 , 4 ]

Now to find the critical values of x

A’(x) = 2 √ 16 – x2 + .5(16 – x2)- .5 (-2x)(2x)

= A’(x) = (2 √ 16 – x2) – 2x2

√ 16 – x2

The function is not defined at x = 4

(2 √ 16 – x2) – 2x2 = 0

√ 16 – x2

Then,

2 ( 16 – x2 ) – 2x2 = 0

32 = 4x2

X = + or – 2 √ 2

So the length of the rectangle is

2x = 2 ( 2 √ 2 ) = 4 √ 2

The width of the rectangle is

Y = √ 16 – x2

Y = √ 16 – ( 2 √ 2 )2

Y = 2 √ 2

Therefore,

A rectangle with a length of 4 √ 2 and width of 2 √ 2 with an area of 16 units can be inscribed in a circle with a radius of 4

A rectangle is inscribed in a circle with a radius of 4.

What is the major equation required for this problem?

Find the critical values of x

What is the largest area the rectangle can have and what are the dimensions?

ANSWERS:

The major equation required for this problem is

The area of the rectangle A = length * width

Since a circle is not a function, we use a semi-circle so that our calculations will be satisfied and more valid.

The equation of this circle is x2 + y2 = 4

So, the coordinates of the rectangle with respect to x are:

Length = 2x width = √ 16 – x2 formula – A(x) = x √ 16 – x2

Thus, the domain for this function would be [ 0 , 4 ]

Now to find the critical values of x

A’(x) = 2 √ 16 – x2 + .5(16 – x2)- .5 (-2x)(2x)

= A’(x) = (2 √ 16 – x2) – 2x2

√ 16 – x2

The function is not defined at x = 4

(2 √ 16 – x2) – 2x2 = 0

√ 16 – x2

Then,

2 ( 16 – x2 ) – 2x2 = 0

32 = 4x2

X = + or – 2 √ 2

So the length of the rectangle is

2x = 2 ( 2 √ 2 ) = 4 √ 2

The width of the rectangle is

Y = √ 16 – x2

Y = √ 16 – ( 2 √ 2 )2

Y = 2 √ 2

Therefore,

A rectangle with a length of 4 √ 2 and width of 2 √ 2 with an area of 16 units can be inscribed in a circle with a radius of 4

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