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# Step-by-Step Math Answer

Use the equation obtained in 1c.

From 1c, the product
*
P
*
=
*
x
*
(110 –
*
x
*
) = 110
*
x
*
–
*
x
*
^{
2
}
, where 0
*
x
*
110.

Find the domain of
*
P
*
.

Since
*
x
*
must be positive, the domain of
*
P
*
is (0, 110).

What should we do next?

Find the asymptotes.

Check for symmetry.

Find the intercepts.

Since 0 is not in the domain, there is no
*
y
*
–intercept. However, there is an open hole at the point (0,
*
P
*
(0)) = (0, 0).

Find the
*
x
*
–intercepts.

To find the
*
x
*
–intercepts we need to solve the equation
*
P
*
= 0. The solutions to this equation are
*
x
*
= 0, 110. Since these are not in the domain there are no
*
x
*
–intercepts. The graph has open holes at the points (0, 0) and (110, 0).

Take the derivative.

Taking the derivative, we find that the derivative function is
*
P
*
′ (
*
x
*
) = 110 – 2
*
x
*
.

Find the points with a horizontal tangent line.

Solving the equation
*
P
*
′ (
*
x
*
) = 0, we find that there is a horizontal tangent line to graph at the point (55,
*
P
*
(55)) = (55, 3025).

Take the second derivative.

Differentiating again, we find that
*
P
*
″ (
*
x
*
) = –2.

Use the second derivative to test the point with a horizontal tangent.

Since
*
P
*
″ (
*
x
*
)
0 for all
*
x
*
, the point (55, 3025) is a relative maximum.

Check for inflection points.

Since the concavity of
*
P
*
is the same for all
*
x
*
in the domain, there are no inflection points.

Examine the graph to determine what value of
*
x
*
maximizes the product.

Looking at the graph, we see that the product is at its highest value when
*
x
*
is 55.

Find the second number,
*
y
*
.

We assumed that the unknown numbers are
*
x
*
and
*
y
*
.

Use the secondary equation
*
x
*
+
*
y
*
= 110. So,

*
y
*
= 110 – 55

= 55.