Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

You work at a farm that wants to build three identical rectangular corrals in a row against the side of a barn using 112 ft of fencing. No fence is needed next to the barn itself. What dimensions (for each individual corral) will maximize the corral areas

Sagot :

Answer:

x  =  14 ft

y  =  56 ft

A(max corral)  =  784/3   =  261.33 ft²

Step-by-step explanation:

Let´s call y the side parallel to the barn, then only one y is going to be fenced.

If we are going to divide the area in three identical corrals we need 4 times x ( the other side perpendicular to the barn)

The perimeter of the rectangular area ( divide in three identical corrals)

112  =  y  +  4*x        or    y  =  112  -  4*x

A (r)  =  x*y

Area as a function of x  is

A(x) = x* ( 112  -  4*x)        A(x)   =  112*x  -  4*x²

Tacking derivatives on both sides of the equation

A´(x)  =  112  - 8*x       A´(x) = 0    112  -  8*x  =  0

x  =  112/8

x  =  14 ft

And  y  =  112  -  4*x     y  =  112 - 56

y  =  56 ft

A(max)  =  14 * 56  =  784 ft²

A(max corral)  =  784/3   =  261.33 ft²

How do we know the area is maximum, tacking the second derivative

A´´(x)  =  - 8     A´´(x) < 0

Then the function A(x) has a maximum at the point x = 14

Kati19

Answer:

14ft by 18 2/3 ft

Step-by-step explanation:

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.