I need to find a maximum area of trapezoid inscribed in circle. The base is diameter of the circle, (1,3) is center of the circle and radius is 4.

Question

I need solution with using derivative.

Answer 1

Normally I would try a video answer, but those have been having problems for the last week. So,

The equation of the circle is (x - 1)² + (y - 3)² = 16.

Let's place the trapezoid in the upper semi-circle. Solve the above for y in terms of x to get

f(x) = 3 + √(16 - (x-1)²)

Now let's see if we can come up with a function for the area of the trapezoid. We will use the above function to determine its height. We can let x take on values between 1 and 5 to represent points along a horizontal radius of the circle that will map to the upper right hand corner of the trapezoid. This might be hard to visualize so a link to a Desmos graph appears at the end of this answer.

So for a given x, the value (x+1) plugged into f(x) will give the y coordinate on the circle of the upper right hand corner of the vertex of the trapezoid. the height of the trapezoid will then be f(x+1) - 3.

The other base of the trapezoid will have a length of (x+1) - (1 - x) = 2x

So the area of the trapezoid is given by A(x) = 1/2 [(f(x+1) - 3] ( 2x + 8), that is 1/2 the height times the sum of the two bases.

Since f(x+1) - 3 = √(16-x²) , A(x) = 1/2 (2x+8) √(16-x²). Find A'(x), using product rule, set equal to zero and find the critical numbers.

Here is the link to the Desmos graph:

desmos.com/calculator/s5orfyeome

You can select that link and choose "go to...".

For A'(x) you should get something like:

-x(x+4)/√(16-x²) + √(16-x²)

There for sure are other ways to set up this problem.When you are on Desmos take a look at the table with values of x and A(x). This will be a way for you to check your answer.

Answer 2

(1,3) appears irrelevant.Clearly, this will be an equilateral trapezoid. Then, if the radius is 4, the diameter is 8. Also, letting the 2 ends be at 0° and 180°, the other two vertices will be at x° and 180 - x°; thus, the length of the parallel side will be 8 cos x°; the height of the trapezoid will be 4 sin xº; then, the area of the trapezoid will be 4(1 + cos xº)4 sin xº = 16(sin xº + sin xº cos xº)

Then, we maximize by taking the derivative and setting it equal to 0.

16 (cos xº + cos² xº - sin² xº) = 16(cos xº + 2 cos² xº - 1)

In parentheses, we see a quadratic; let y = cos xº; then 2y² + y - 1 = 0

This factors as (2y - 1)(y + 1) We don't need to consider y = -1 (our construction tells us that we only need to consider positive values of cos). Thus, y = 1/2, or cos x = 1/2; then sin x = √3/2

The maximum area is 4(1 + 1/2)4 √3/2 = 12√3 Note that this is less than the area o the semicircle, 8 π