Homework Help: intermediate algebra

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

A farmer with 3000 feet wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?

intermediate algebra - drwls, Tuesday, March 9, 2010 at 8:41am
Let 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 9:11pm
where 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 9:11pm
where 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 9:36am
Let 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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