Help with math... Quadratics?

Question

18. Bill has 75 feet of fencing to build a pen for his goats. He will use the side of his barn for one side of the
pen.
a) What should the dimensions of the pen be to maximize the area of the pen?
b) What is the maximum area?

Forgot how to do this.

Answer 1

If the pen can be any shape, a semicircle would maximize the area. If the barn is the straight side, and the fencing is the curve, then the cirsumfrence of the full circle is 150 ft. C = pi X d gives us d = 47.7 ft. The area of the semicircle would be (pi r^2)/2. the area of the pen is 995 ft^2

If the pen has to be a quadrilateral, then a rectangle with the barn as one of the long sides gives maximun area. If the long side is half of the fencing that would give maximum area. The dimentions are 37.5 ft X 18.75 ft. The area is 703.125 ft^2.

Answer 2

Little Fairy is incorrect because one side of the barn is used as a side. So if the sides perpendicular to the barn are 'a' and the side parallel to barn is 'b'. Then 2a + b = 75...Area=ab. According to the first formula, b = 75-2a. plug that in for 'b' in the Area formula and get Area=a(75-2a) or -2a^2+75a. Plug that into a calculator and solve for the max. That is your maximum 'a' value. Plug that value in for a to solve for 'b' in 2a + b = 75. Those are your dimensions. Multiply them together and you have your maximum area.

Answer 3

i think maximum area is 485 ft

Answer 4

if the sides of the pen are a and b, then:
2a+2b=75 ===> a+b=37.5 ====> a=37.5-b
a*b=(37.5-b)*b=37.5b-b^2
37.5-2b=0
b=37.5/2=18.75 ft
=> a=18.75 ft
area= a*b=351.56 sq feet