Chapter 11.1, Problem 6E

(a)

To determine

To draw:a diagram that shows the base $x$ and height $y$ of the triangle.

Answer to Problem 6E

  

Explanation of Solution

Given:

Right triangle with a hypotenuse of $\sqrt{18}$ meters

Calculation:

Here, draw a triangle with the specified hypotenuse length of V18, a base $lengthx,andaheighty$ . In order to make Part B more clear, re-express $y$ in terms of $x$ , derived by using the Pythagorean

Theorem:

  $\begin{array}{l}{a}^2+b^2=c^2\\ x^2+y^2=18\\ y^2=18-x^2\\ y=\sqrt{18-x^2}\end{array}$

  

Conclusion:

Thus, the required diagram is drawn.

(b)

To determine

To verify:the area $A$ of the triangle is $A=\frac{1}{2}x\sqrt{18-x^2}$ .

Explanation of Solution

Given:

  $A=\frac{1}{2}x\sqrt{18-x^2}$

Proof:

The area of a triangle is given by $\frac{1}{2}b\cdot h$ , so here compute this with our triangle’s base length of $x$ and height of $\sqrt{18-x^2}$ (found in Part A)

  $A=\frac{1}{2}b\cdot h=\frac{1}{2}x\sqrt{18-x^2}$

Conclusion:

Thus,the area $A$ of the triangle is given by

(c)

To determine

To complete:the table,observe the behavior and find the limit.

Answer to Problem 6E

  $\underset{x\to 3}{\lim }A=4.5$

Explanation of Solution

Given:

  $\underset{x\to 3}{\lim }A$

Calculation:

$x$	2	2.5	2.9	3	3.1	3.5	4
$A$	3.74	4.28	4.49	4.50	4.49	4.20	2.83

From the table above, the function appears to take values close to $4.5asxapproaches3$ from both the left and right, thus, $\underset{x\to 3}{\lim }A=4.5$

Conclusion:

The value of the limit is$\underset{x\to 3}{\lim }A=4.5$

(d)

To determine

To graph:the areas functions and verify that the area is maximum when $x=3$ meters.

Answer to Problem 6E

  

Explanation of Solution

Proof:

Graphing this function, it takes its highest value when $x=3$

  

Conclusion:

Thus, the area is maximum when $x=3$ meters