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April 20, 2014

April 20, 2014

Posted by **irma** on Monday, March 8, 2010 at 11:30pm.

- intermediate algebra -
**drwls**, Tuesday, March 9, 2010 at 8:41amLet x be the side length parallel to the highway. The side lengths perpendicular to the highway must be

(1/2) (3000 -x)

The area is

A = (x/2)*(3000-x) = 3000x -x^2/2

When there is a maximum,

dA/dx = 0 = 3000 - 2x

x = 1500 feet

I used calculus but you can try completeing the square or try various values of x until you get a maximum area.

The enclosed area will be 1500 x 750 = 1,125,000 ft^2

- intermediate algebra -
**joe**, Wednesday, February 2, 2011 at 9:11pmwhere did you get the 750? I have a similar problem here:

a farmer wants to build a rectangular pen using a side of a barn and 60ft of fence. find the dimensions and area of the largest such pen

- intermediate algebra -
**joe**, Wednesday, February 2, 2011 at 9:11pmwhere did you get the 750? I have a similar problem here:

a farmer wants to build a rectangular pen using a side of a barn and 60ft of fence. find the dimensions and area of the largest such pen

- intermediate algebra -
**prince**, Sunday, February 17, 2013 at 9:36amLet l = measure of the parallel side of the highway in meters

w = measure of the perpendicular side of the highway in meters

l + w = 3000

l = 3000 - w

length = 3000 - w

width = w

A= lw

**since we will only use 1 side of the length, we will use:

A= [(3000-w)/2]w 0r w[(3000-w)/2]

= -(w^2)/2 + 1500w

**complete the square

= -1/2 (w^2 - 3000w + 225,000) + 1,125,000

= -1/2 (w-1500)^2 + 1,125,000

w=1500

A max.= 1,125,000

**substitution

l=(3000-w)/2

=(3000-1500)/2

=750

dimensions: 1500 x 750

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