Solve this question using derivatives please. A farmer has 750 m of fencing. The farmer wants to enclose a rectangular area on all four sides, and then divide it into four pens of equal size with the fencing parallel to one side of the rectangle. What is the largest possible area of each of the four pens?

Step-by-step explanation:

The area of one pen will be largest when the area of a ractangle is the largest.

x - one side of rectangular area      (x>0)

y - second side of rectangular area   (y>0)

The farmer has 750 m of fencing, so if we subtract the sides x and fencing used for dividing into pens we receive:

y = 750 - 5x          (750-5x>0  ⇒  x<150)

Area of the rectangle:  A = x•y

A(x) = x•(750 - 5x)             D=(0, 150)

A(x) = -5x² + 750x            

A'(x) = -5•2x + 750•1 = -10x + 750 = -10(x - 75)

if x=75 then A'(x) = 0

so x = 75 is critical point

First Derivative Test:

A'(x) > 0  ⇔ -10(x-75)>0 ⇔  x-75<0  ⇔ x<75

A'(x) < 0  ⇔ -10(x-75)<0 ⇔  x-75>0  ⇔ x>75

A′(x)>0 to the left of x=75 and A′(x)>0 to the right of x=75 then x=75 is a maximum. {rational but also global in the interval (0,150)}

A(75) = 75•(750 - 5•75) = 75•375 = 28125 m²

So the largest possible area of each of the four pens is:  

8125 : 4 = 7031.25 m²

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00