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Q. 11

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Found in: Page 120

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# A wire $10$ meters long is to be cut into two pieces. One piece will be shaped as a square, and the other piece will be shaped as a circle. See the figure.Part (a): Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the square.Part (b): What is the domain of A?Part (c): Graph $A=A\left(x\right)$. For what value of x is A smallest?

Part (a): Total area A enclosed by the pieces of wire as a function is $\mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+\frac{25-20\mathrm{x}+4{\mathrm{x}}^{2}}{\mathrm{\pi }}$.

Part (b): The domain of A is $\left(0,\frac{5}{2}\right)$.

Part (c): On plotting the function, we get,

The value of x for which A is smallest at $x\approx 1.4$m.

See the step by step solution

## Part (a) Step 1. Given information.

Assume x m to be the length of side of the square.

Perimeter$\left(p\right)=4x$

From the given figure,

Circumference of the circle formed is $10-4x$m.

Assume r(x) to be the radius of the circle formed. Then it can be written,

localid="1645810318472" $2\mathrm{\pi r}\left(\mathrm{x}\right)=10-4\mathrm{x}\phantom{\rule{0ex}{0ex}}\mathrm{r}\left(\mathrm{x}\right)=\frac{10-4\mathrm{x}}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}\mathrm{r}\left(\mathrm{x}\right)=\frac{5-2\mathrm{x}}{\mathrm{\pi }}$

## Part (a) Step 2. Calculate the total area.

Consider the given question,

$A\left(x\right)={x}^{2}+{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+\mathrm{\pi }{\left(\frac{5-2\mathrm{x}}{\mathrm{\pi }}\right)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{A}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+\frac{25-20\mathrm{x}+4{\mathrm{x}}^{2}}{\mathrm{\pi }}$

## Part (b) Step 1. Find the domain of A.

Consider the given question,

$\mathrm{r}\left(\mathrm{x}\right)=\frac{5-2\mathrm{x}}{\mathrm{\pi }}$. Also $r\left(x\right)>0$. Then,

$\frac{5-2\mathrm{x}}{\mathrm{\pi }}>0\phantom{\rule{0ex}{0ex}}5-2x>0\phantom{\rule{0ex}{0ex}}x<\frac{5}{2}$

x cannot be negative. As length cannot be negative.

Therefore, the domain is $\left(0,\frac{5}{2}\right)$.

## Part (c) Step 1. Plot the function.

On plotting the function, we get,

From the graph, we can say that A is smallest at $x\approx 1.4$m.