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# Step-by-Step Math Answer

Find the primary equation.

The strength has to be maximized, so the primary equation is
*
S
*
=
*
kh
*
^{
2
}
*
w
*
, where
*
k
*
is the constant of proportionality.

Find the secondary equation.

Using the Pythagorean Theorem, we find that the secondary equation is

*
w
*
^{
2
}
+
*
h
*
^{
2
}
= (24)
^{
2
}
= 576.

Solve the secondary equation for
*
h
*
^{
2
}
.

Solving the secondary equation for
*
h
*
^{
2
}
, we find that
*
h
*
^{
2
}
= 576 –
*
w
*
^{
2
}
.

Express the primary equation in terms of
*
w
*
.

In terms of
*
w
*
, we can express the primary equation as

*
S
*
=
*
kw
*
(576 –
*
w
*
^{
2
}
) = 576
*
kw
*
–
*
kw
*
^{
3
}
, where 0
*
w
*
24.

Differentiate
*
S
*
.

Differentiating
*
S
*
, we find that
*
S
*
′ (
*
w
*
) = 576
*
k
*
– 3
*
kw
*
^{
2
}
.

Find the critical numbers.

Solving the equation
*
S
*
′ (
*
w
*
) = 0, we find that the solutions are

However, negative values of
*
w
*
aren't allowed, so the critical number is

What should we do next?

Find the value of
*
h
*
.

Verify that
*
S
*
is maximized for this value of
*
w
*
.

Let's verify that
*
S
*
is maximized for this value of
*
w
*
. Begin by finding the second derivative. Differentiating again, we find that
*
S
*
″ (
*
w
*
) = –6
*
kw
*
.

Apply the Second Derivative Test.

Since
*
S
*
″ (
*
w
*
)
0 at the critical number, the Second Derivative Test tells us that
*
S
*
is maximized when

Find the value of
*
h
*
.

We have found that
*
S
*
is maximized when

Using the secondary equation, we find that the corresponding value of
*
h
*
is