ECON 331
Two Variable Optimization
Using Calculus For Maximization Problems
One Variable Case
If we have the following function
y
= 10
x
−
x
2
we have an example of a
dome shaped
function.
To
fi
nd the maximum of the dome, we
simply need to
fi
nd the point where the slope of the dome is zero, or
dy
dx
= 10
−
2
x
= 0
10 = 2
x
x
= 5
and
y
= 25
Two Variable Case
Suppose we want to maximize the following function
z
=
f
(
x, y
) = 10
x
+ 10
y
+
xy
−
x
2
−
y
2
Note that there are two unknowns that must be solved for:
x
and
y
. This function is an
example of a
threedimensional dome.
(i.e. the roof of
BC Place
)
To solve this maximization problem we use
partial derivatives.
We take a partial
derivative for each of the unknown choice variables and set them equal to zero
∂
z
∂
x
=
f
x
= 10 +
y
−
2
x
= 0
The slope in the ”x” direction = 0
∂
z
∂
y
=
f
y
= 10 +
x
−
2
y
= 0
The slope in the ”y” direction = 0
This gives us a set of equations, one equation for each of the unknown variables. When
you have the same number of independent equations as unknowns, you can solve for each of
the unknowns.
rewrite each equation as
y
= 2
x
−
10
x
= 2
y
−
10
substitute one into the other
x
= 2(2
x
−
10)
−
10
x
= 4
x
−
30
3
x
= 30
1
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x
= 10
similarly,
y
= 10
REMEMBER:
To maximize (minimize) a function of many variables you use the
technique of partial di
ff
erentiation. This produces a set of equations, one equation for each
of the unknowns. You then solve the set of equations simulaneously to derive solutions for
each of the unknowns.
Second order Conditions (second derivative Test)
To test for a maximum or minimum we need to check the second partial derivatives. Since
we have two
fi
rst partial derivative equations (
f
x
,
f
y
) and two variable in each equation, we
will get four
second partials (
f
xx
, f
yy
, f
xy
, f
yx
)
Using our original
fi
rst order equations and taking the partial derivatives for each of them
(a second time) yields:
f
x
= 10 +
y
−
2
x
= 0
f
y
= 10 +
x
−
2
y
= 0
f
xx
=
−
2
f
yy
=
−
2
f
xy
= 1
f
yx
= 1
The two partials,
f
xx
, and
f
yy
are the direct e
ff
ects of of a small change in
x
and
y
on
the respective slopes in in the
x
and
y
direction. The partials,
f
xy
and
f
yx
are the indirect
e
ff
ects, or the cross e
ff
ects of one variable on the slope in the other variable’s direction. For
both
Maximums and Minimums
, the direct e
ff
ects must outweigh the cross e
ff
ects
Rules for two variable Maximums and Minimums
1. Maximum
f
xx
<
0
f
yy
<
0
f
yy
f
xx
−
f
xy
f
yx
>
0
2. Minimum
f
xx
>
0
f
yy
>
0
f
yy
f
xx
−
f
xy
f
yx
>
0
3. Otherwise, we have a
Saddle Point
From our second order conditions, above,
f
xx
=
−
2
<
0
f
yy
=
−
2
<
0
f
xy
= 1
f
yx
= 1
and
f
yy
f
xx
−
f
xy
f
yx
= (
−
2)(
−
2)
−
(1)(1) = 3
>
0
therefore we have a maximum.
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 Spring '09
 KelvinKwainger

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