Graphics Programs Reference
In-Depth Information
storage space for 4,000 bushels. Each acre yields an average of 110 bushels
of wheat or 30 bushels of barley. If the net profit per bushel of wheat (after
all expenses have been subtracted) is $1.30 and for barley is $2.00, how
should the farmer plant the 75 acres to maximize profit?
We begin by formulating the problem mathematically. First we express
the objective, that is, the profit, and the constraints algebraically, then we
graph them, and lastly we arrive at the solution by graphical inspection and
a minor arithmetic calculation.
Let
x
denote the number of acres allotted to wheat and
y
the number of
acres allotted to barley. Then the expression to be maximized, that is, the
profit, is clearly
P
=
(110)(1
.
30)
x
+
(30)(2
.
00)
y
=
143
x
+
60
y
.
There are three constraint inequalities, specified by the limits on expenses,
storage, and acreage. They are respectively
120
x
+
210
y
≤
15
,
000
110
x
+
30
y
≤
4
,
000
x
+
y
≤
75
.
Strictly speaking there are two more constraint inequalities forced by the
fact that the farmer cannot plant a negative number of acres, namely,
x
≥
0
,
y
≥
0
.
Next we graph the regions specified by the constraints. The last two say
that we only need to consider the first quadrant in the
x
-
y
plane. Here's a
graph delineating the triangular region in the first quadrant determined by
the first inequality.
X = 0:125;
Y1 = (15000 - 120.*X)./210;
area(X, Y1)
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