SOLUTION: The figure shows an isosceles triangle with equal sides of length "a" surmounted by a semicircle. What should the measure of angle θ be in order to maximize the total area?

Algebra ->  Surface-area -> SOLUTION: The figure shows an isosceles triangle with equal sides of length "a" surmounted by a semicircle. What should the measure of angle θ be in order to maximize the total area?       Log On


   



Question 1186062: The figure shows an isosceles triangle with equal sides of
length "a" surmounted by a semicircle. What should the measure of angle θ be in order to maximize the total area?
https://imgur.com/a/tyyqfL4

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!


Refer to the figure above, where we've sliced the original picture down the middle. Here alpha+=+theta%2F2+. If we maximize this figure in terms of alpha, we just need to double that to find theta.
A%5Barc%5D = +%281%2F4%29%2Api%2Ar%5E2+
A%5Btriangle%5D = +%281%2F2%29r%2Ah+
Observe:
+r+=+a%2Asin%28alpha%29+
+h+=+a%2Acos%28alpha%29+
Substituting for r and h:
A%5Barc%5D = +%281%2F4%29%2Api%2Aa%2Asin%5E2%28alpha%29+
A%5Btriangle%5D = +%281%2F2%29%28a%2Asin%28alpha%29%29%2A%28a%2Acos%28alpha%29%29+
+A%5Btotal%5D+ = A%5Barc%5D+%2B+A%5Btriangle%5D+
= +%281%2F4%29pi%2Aa%5E2%2Asin%5E2%28alpha%29+ + +%281%2F2%29%28a%2Asin%28alpha%29%29%2A%28a%2Acos%28alpha%29%29+
...factoring out +%281%2F4%29a%5E2%2Asin%28alpha%29+
= (*)
Now take the derivative of A%5Btotal%5D with respect to +alpha+:
+dA%5Btotal%5D+%2Fd%5Balpha%5D+ =

= +%281%2F4%29a%5E2%2A%28pi%2Asin%282%2Aalpha%29%2B2cos%282%2Aalpha%29%29+
... substitute theta+=+2%2Aalpha+ ...
= +%281%2F4%29a%5E2%28pi%2Asin%28theta%29%2B2cos%28theta%29%29+

This last function is zero at approx. theta+ = 2.575rad, theta= 5.716rad, ...
By graphing, one can see 2.575rad results in a maximum area, while 5.716 results in a minimal area.
2.575rad is approx 147.54degrees.
Ans: theta = 147.54%5Eo maximizes the total area.
Just in case...
If you were interested in the value of the total maximal area, note that you would need a modified form of (*) because that equation computes 1/2 of the total area:
+A%5Btotal%5D+ =