Q. 5 - Calculus

Question

Consider the graph of the function f shown next. Define A(x) to be the area of the region between the graph of f and the x-axis from 0 to x. We will count areas of regions above
the x-axis positively and areas of regions below the x-axis negatively.

Show that your answers for the local and global extrema of A(x) are reasonable by using optimization techniques on the area function $A\left(x\right)=3x^4-32x^3+90x^2$?

Step-by-Step Solution

Answer

The global maxima of A(x) is $A\left(3\right)=189$ and the global minima is $A\left(0\right)=0$

Step 1. Given Information.

The graph.

The area function:

$A\left(x\right)=3x^4-32x^3+90x^2$

Step 2. Find the critical points.

Differentiate with respect to x,

$$\begin{gathered} A\left(x\right)=3x^4-32x^3+90x^2 \\ A\text{'}\left(x\right)=12x^3-96x^2+180x \\ (x^2-8x+15)12x=0 \\ x(x-5)(x-3)=0 \\ x=0,3,5 \end{gathered}$$

Then $x=0,3$ is the local minimum and $x=5$ is the local maximum.

Step 3. Values of A(x).

The value of A(x) are:

$$\begin{gathered} A\left(0\right)=0 \\ A\left(1\right)=61 \\ A\left(2\right)=152 \\ A\left(3\right)=189 \\ A\left(4\right)=160 \\ A\left(5\right)=125 \end{gathered}$$

The global maxima of A(x) is $A\left(3\right)=189$ and the global minima is $A\left(0\right)=0$