Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come to the conclusion, using the pythagorean theorem, that the height of the trapezoid is h=1−x24−−−−−√h=1−x24
The formula for the area of a trapezoid is A=a+b2(h)A=a+b2(h) I am basically solving for the aa here and so far I have:
A=x+22(1−x24−−−−−√)A=x+22(1−x24)
I simplified this in order to be able to easily take the derivative of it as such:
A=x+22(4−x24−−−−−−√)A=x+22(4−x24)A=x+22(4−x2−−−−−√2)A=x+22(4−x22)A=(14)(x+2)(4−x2−−−−−√)A=(14)(x+2)(4−x2)Then I took the derivative:
A′=14[(1)(4−x2−−−−−√)+(x+2)((12)(4−x2)−1/2(−2x)]A′=14[(1)(4−x2)+(x+2)((12)(4−x2)−1/2(−2x)]This all simplified to:
A′=14[4−x2−−−−−√−2x2−4x4−x2−−−−−√]A′=14[4−x2−2x2−4x4−x2]Next, I set the derivative equal to zero in order to find the maximum point (I know that I can prove that it is actually a max point by taking the second derivative later)
14[4−x2−−−−−√−2x2−4x4−x2−−−−−√]=014[4−x2−2x2−4x4−x2]=04−x2−−−−−√−2x2−4x4−x2−−−−−√=04−x2−2x2−4x4−x2=04−x2−2x2−4x4−x2−−−−−√=04−x2−2x2−4x4−x2=04−x2−2x2−4x=04−x2−2x2−4x=0−3x2−4x+4=0−3x2−4x+4=0so I got:
x=−2orx=23x=−2orx=23x=−2x=−2 wouldn't make sense so I chose x=23x=23 and plugged it back into my original formula for the area of this trapezoid but my answer doesn't seem to match any of the multiple choice solutions. Where did I go wrong?
calculus optimization