Chapter 1.7, Problem 20E

Solution

To write: The specified quantity as a function of specified variable.

The required function is $f\left(x\right)=\frac{24x-x^3}{8}$ , $f\left(x\right)$

Given data: The isosceles triangle has its base along x -axis with one base vertex at the origin and its vertex at the origin and vertex on the curve $y=6-x^2$ . The area of the triangle is a function length of the base.

Method/Formula used:

The area A of an isosceles triangle is $A=\frac{1}{2}\text{base×height}$ .

Calculation:

The isosceles triangle with its base OB along x -axis with one base vertex at the origin O and other vertex at point A on the curve $y=6-x^2$ is shown in Fig. (1).

  

The area of the triangle is a function length of the base.

Let the co-ordinates of the vertex A is $\left(x,y\right)$ and M be mid-point of the base OB . Then, as the median of an isosceles triangle is perpendicular to the base, AM is the height of the triangle OAB . Therefore, the area S of the triangle OAB is,

  $S=\frac{1}{2}OB\times AM$ .

Write $OB=b$ , and $AM=h$ , then

  $S=\frac{1}{2}b\times h$ … (1)

From the Fig. (1), $OM=b/2$ , therefore, the coordinates of point B are $\left(b,0\right)$ and that of point A are $\left(b/2,h\right)$ because $\left(b/2,h\right)$ . Since the point A lies on the curve $y=6-x^2$ , therefore, it must satisfy the equation of the curve, i.e.,

  $\begin{array}{c}h=6-{\left(\frac{b}{2}\right)}^2\\ =6-\frac{b^2}{4}\\ =\frac{24-b^2}{4}\end{array}$ … (2)

Substitute, $\frac{24-b^2}{4}$ for h in equation (1),

  $\begin{array}{c}S=\frac{1}{2}b\left(\frac{24-b^2}{4}\right)\\ =\frac{24b-b^3}{8}\end{array}$

Here, b is the base length.

Thus, the area of the triangle is a function length of the base is $S\left(b\right)=\frac{24b-b^3}{8}$ .