SOLUTION: a farmer has 500 feet of fencing with which to build a rectangular livestock pen and wants to enclose the maximum area.
use a variable to label length and width of the rectangle
Algebra.Com
Question 186379: a farmer has 500 feet of fencing with which to build a rectangular livestock pen and wants to enclose the maximum area.
use a variable to label length and width of the rectangle
find a variable expression for the area of the pen
this is totally killing me can someone please help??? thank you
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
a farmer has 500 feet of fencing with which to build a rectangular livestock
pen and wants to enclose the maximum area.
:
use a variable to label length and width of the rectangle
find a variable expression for the area of the pen
:
Let x = width of the pen
Let L = length of the pen
:
write an equation for the total fencing available (the perimeter)
2L + 2x = 500
Simplify, divide equation by 2
L + x = 250
L = (250-x)
:
The area = L * x, substitute (250-x) for x
A = x(250-x)
:
A = -x^2 + 250x; this is the variable expression for the area of the pen
:
You can calculate the max area using the axis of symmetry formula x=-b/(2a)
In this problem:
x = -250/(2*-1)
x = -250/-2
x = 125 is the width for max area
Note that would be a square: L = 250-x, right?
L = 250 - 125
L = 125 also
:
To find the actual area: substitute 125 for x in the equation
A = -(125^2) + 250(125)
A = -15625 + 31250
A = + 15625 sq/ft is the max area
:
Note this corresponds to a square field: 125 * 125 = 15625 sq/ft
:
To graph this equation, area = y; write it:
y = -x^2 = 250x
On a TI83 or similar just enter y= -x^2 +25x, scale it x:-100,+300; y:-2000,+16000
;
it should look like this
:
:
Note the max area (y) occurs when x = 125
:
If you have to plot this on graph paper, you have to make up an x/y table
and plot about a dozen points from x=0 to x=+250. let me know if you have
to do that and don't know how to do it. Carl
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