Posted by **angel** on Wednesday, April 18, 2012 at 8:44pm.

A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 80 feet of fence? What should the dimensions of the garden be to give the area?

- word problem -
**Reiny**, Wednesday, April 18, 2012 at 9:44pm
He will need one length and two widths

let the length by y ft

let the width be x ft

2x + y = 80

y = 80-2x

area = xy

= x(80-2x)

= -2x^2 + 80x

The x value to produce the maximum area is the x of the vertex of this parabola.

If you know Calculus

d(area)/dx = -4x + 80= 0 for a max of area

x = 20

then y = 80-2x = 40

the field should be 20 by 40 , with 40 being the one side.

non-Calculus:

the x of the vertex of y = ax^2 + bx +c is -b/(2a)

which in this case is -80/(2(-2)) = 20

continue as above.

or you can complete the square:

area = -2(x^2 - 40x + **400 - 400 **)

= -2(x-20)^2 + 800

max of area is 800 when x = 20 and y = 40

- word problem -
**gordana**, Sunday, September 14, 2014 at 9:47am
The farmer wants to use the 500 m to an enclosure divided into two equal areas. What is the total maximum area that can is achieved with the 500 m fence. use differential calculus to the solution.

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