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December 20, 2014

December 20, 2014

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

- word problem -
**Reiny**, Wednesday, April 18, 2012 at 9:44pmHe 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 9:47amThe 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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