THe director of the zoo wants to use 170m of fencing to enclose a petting area of 1750m^2. What are the dime?

The director of the zoo wants to use 170m of fencing to enclose a petting area of 1750m^2. What are the dimensions of the petting zoo?|||Do you know it has to be rectangular? I%26#039;ll assume so. If you let x=length of one side, then you can figure out the length of the other side (temporarily call that y) since we know the perimeter of the rectangle.





2x + 2y = 170


2y = 170 - 2x


y = 85 - x





The area enclosed is 1750 m^2, so


x y = 1750


x (85 - x) = 1750


85x - x^2 = 1750


x^2 - 85x + 1750 = 0


(x-35) (x-50) = 0





x=35 or x=50





Either solution will work for the length of one side. y will turn out to be the other.





Dimensions are 35m by 50m|||There are multiple answers to this, depending on the shape of the pen.


If you go with the rectangle, you need to solve the simultaneous equations:


2a+2b = 170, (ie a=+b = 85)


ab = 1750





so a = 85-b


substituting into the second equation gives


85b-b.b =1750


b.b-85b+1750=0


(b-35)(b-50)=0





so the dimensions are 35m x 50m.





But the most efficient use of fencing would be to make a circle.