Question

topic 2 composing and decomposing functions
skills practice continued
problem set 3: exploring cubic functions
complete each table. include an expression for the volume. circle the relative maximum or
minimum if there is one.
1

Explanation:

Step1: Analyze the volume formula

The volume formula for the box is \(V = h(8 - 2h)(10 - 2h)\). Expand this formula:

$$
LATEXBLOCK0
$$

Step2: Find the derivative of the volume function

Let \(V(h)=4h^{3}-36h^{2}+80h\). The derivative \(V^{\prime}(h)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\) is:

$$
LATEXBLOCK1
$$

Step3: Set the derivative equal to zero and solve for \(h\)

Set \(V^{\prime}(h)=0\), so \(12h^{2}-72h + 80=0\). Divide through by \(4\): \(3h^{2}-18h + 20=0\).
Using the quadratic formula \(h=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) where \(a = 3\), \(b=-18\), \(c = 20\)

$$
LATEXBLOCK2
$$

We get \(h_1\approx3 + 1.53=4.53\) (but when \(h = 4\), \(8-2h=0\), so \(h = 4.53\) is not in the domain \(0\lt h\lt4\)) and \(h_2\approx3-1.53 = 1.47\approx1.5\)

Step4: Check the second - derivative

The second - derivative \(V^{\prime\prime}(h)=24h-72\). When \(h = 1.5\), \(V^{\prime\prime}(1.5)=24\times1.5-72=36 - 72=-36\lt0\)

Answer:

The value \(h = 1.5\) is a relative maximum.