Calculus Problem

Question

A building is to be built and will take the form of a semicircular shape and will have a rectangular doorway. The building will be 6 m long and the door must be at least 1 m vertically from the roof of the building.

A) What is the largest area the doorway can have and what are its dimensions(give the area to the nearest 0.01 m^2 and the dimensions to the nearest cm)

Answer 1

Draw a rectangular door with the center of the length of the door on top of th center of the diameter of the semicircle. The height of the corner of the door is 1 away vertically from the roof. If we call the height of the door x, then the length  of the segment from the diameter to the roof (along the side of the door, past the top of the door and ending at the roof) can be represented as x+1. The Point where the segment hits the roof to the center of diameter is 3 since it is a radius. Use Pythagorean Theorem to get half the width of the door to be sqrt (3^2 - (x +1)^2) which is sqrt(9-x^2-2x-1) or sqrt (8-2x-x^2). So the width of the door is 2*(sqrt 8-2x-x^2). 

So area= x * 2 * sqrt 8-2x-x^2. You can now differentiate using product rule to get A' = 2sqrt8-2x-x^2  +2x(-2-2x) / (8-2x-x^2). Set=0 and simplify to 8-3x-4x^2 = 0 and stick in quadratic formula to get x=1.097 .width = 4.34 and A=4.76 sq m.