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 \(V = h(8 - 2h)(10 - 2h)\). Expanding this:

$$
LATEXBLOCK0
$$

Step2: Find the derivative

The derivative \(V^\prime\) using the power rule \((x^n)^\prime=nx^{n - 1}\) is:

$$ V^\prime=12h^{2}-72h + 80 $$

Set \(V^\prime = 0\) for critical points:

$$12h^{2}-72h + 80=0$$

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

$$
LATEXBLOCK1
$$

We get \(h\approx1.47\) and \(h\approx4.53\). But since \(h\) represents the height of the box and \(8-2h>0\) and \(10 - 2h>0\) (dimensions of length and width must be positive), \(h<4\). So we consider \(h\approx1.47\)

Step3: Use the second - derivative test

The second derivative \(V^{\prime\prime}=24h-72\)
When \(h = 1.47\), \(V^{\prime\prime}(1.47)=24\times1.47-72=35.28-72=- 36.72<0\)

Answer:

The relative maximum occurs at \(h\approx1.5\) (since \(h\approx1.47\approx1.5\) in the context of the table). The volume at \(h = 1.5\) is \(52.5\) in.³ and it is a relative maximum.