A window is to be built in the shape of a semi-circle over a rectangle: {{CMD_1}} {{ENV_2}} {{CMD_3}} If the outer perimeter is {{MATH_4}} feet, maximize the area. Round your answer to the closest tenth of a square foot. {{CMD_5}} Let's label the picture. {{CMD_6}} {{ENV_7}} {{CMD_8}} {{CMD_9}} {{CMD_10}} The outer perimeter, {{MATH_11}}, is given by {{MATH_12}}. Since we know that {{MATH_13}}, we can express {{MATH_14}} in terms of {{MATH_15}}. {{MATH_16}}. The area, A , consists of the area of the rectangle + the area of the semicircle: {{MATH_17}} {{CMD_18}} {{CMD_19}} We can now express the area {{MATH_20}} as a function of {{MATH_21}}: {{MATH_22}}. The domain of A is the closed interval {{MATH_23}}.(Remember: P=10!) {{CMD_24}} {{CMD_25}} Now we have to find the global maximum of {{MATH_26}} on its domain. First, we have to find the critical points of {{MATH_27}}. Since {{MATH_28}}, it follows that the function {{MATH_29}} has its only critical point at {{MATH_30}} . {{CMD_31}} {{CMD_32}} Now, we evaluate {{MATH_33}} at the end points and at the critical point and then compare the values. {{MATH_34}} {{MATH_35}} {{MATH_36}} This seems complicated. It is better to argue that the derivative {{MATH_37}} on {{MATH_38}} and {{MATH_39}} on {{MATH_40}}, since {{MATH_41}}. This implies that the global maximum occurs at the critical point.