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 of the box is \(V(h)=h(8 - 2h)(10 - 2h)\). Expand it:

$$
LATEXBLOCK0
$$

Step2: Find the derivative

Differentiate \(V(h)\) with respect to \(h\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(V^\prime(h)=12h^{2}-72h + 80\).

Step3: Set the derivative equal to zero

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

$$
LATEXBLOCK1
$$

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

Step4: Use the second - derivative test

Differentiate \(V^\prime(h)\) to get \(V^{\prime\prime}(h)=24h-72\). Substitute \(h = 1.5\) into \(V^{\prime\prime}(h)\): \(V^{\prime\prime}(1.5)=24\times1.5-72=36 - 72=-36\lt0\)

Answer:

The volume function \(V(h)=4h^{3}-36h^{2}+80h\) has a relative maximum at \(h = 1.5\) inches. The relative maximum volume is \(52.5\) cubic inches.