Chapter 10
Experimental Design
2
(c)
For a contour level of c, we set f(x,y) = c.
Taking natural logarithms of both sides of the equation
[
]
2
2
2
1
)
5
(
)
2
(
−
+
−
−
y
x
e
= c, we find -
2
1
[(x-2)
2
+(y-5)
2
] = ln(c).
Note that c must be positive (because the
exponential function is always positive) and less than 1 (since the expression in the exponent is always
negative and the exponential function is less than 1 for negative arguments).
Thus, ln(c) must be
negative.
Multiplying through by -2 then yields the familiar equation
(x-2)
2
+(y-5)
2
= -2ln(c), where -
2ln(c) is a positive number.
This is the equation of a circle in the plane with radius equal to the square
root of -2ln(c).
That is, all contours for f(x,y) are circles in the plane.
(d) Sketching the contours should show that the maximum is achieved when the expression in the
exponent of f(x,y) is 0.
Because (x-2)
2
+(y-5)
2
is nonnegative, it can only equal 0 when both x = 2 and
y = 5.
Another method of obtaining the maximum point would be to take partial derivatives of f(x,y),
set them equal to 0, and solve.
For example, the equations
x
∂
∂
[
]
2
2
2
1
)
5
(
)
2
(
−
+
−
−
y
x
e
= -(x-2)
[
]
2
2
2
1
)
5
(
)
2
(
−
+
−
−
y
x
e
= 0
and
y
∂
∂
[
]
2
2
2
1
)
5
(
)
2
(
−
+
−
−
y
x
e
= -(y-5)
[
]
2
2
2
1
)
5
(
)
2
(
−
+
−
−
y
x
e
= 0
have the unique solution x = 2, y =5.
4.
(a) The following surface plot of the function f(x,y) was created in MathCAD.
The surface is an inverted
trough that falls off symmetrically on either side of the line y = x in the x-y plane.
Note from the graph that there
are infinitely many maxim, which all occur on the line y = x.
(b) Pick any value of c between 0 and 1; i.e., 0 < c < 1. To find the contour level for c , set f(x,y) = c, take
natural logarithms of both sides, and solve for y:
-(x-y)
2
= ln(c), or , (x-y)
2
= -ln(c) = k ,