Answer
(a) $A(x)= x^2+\dfrac{(5-2x)^2}{\pi}$
(b) domain $(0,2.5)$
(c) See graph below. $x\approx1.40$
Work Step by Step
(a) The wire whose length is $10-4x$ meters will be used to form a circle. This means that the circumference of the circle is $10-4x$ meters.
Recall that the circumference $C$ of a circle is given by the formula $C=2\pi{r}$ where $r$ is the radius of the circle.
Since $C=2\pi{r}$ and $C=10-4\pi$, then $2\pi{r}=10-4x$.
Solve for $r$ to obtain:
$$\begin{align*}
2\pi{r}&=10-4x\\
r&=\frac{10-4x}{2\pi}\\
r&=\frac{5-2x}{\pi}\\
\end{align*}$$
The area of a circle is $\pi{r^2}$ square units where $ r$ is the radius, while the area of a square is $s^2$ square units where $s$ is the length of each side.
Thus, the total area $A$ is:
$$\begin{align*}
A(x)&=\text{area of the square} + \text{area of the circle}\\
&=x^2+\pi r^2\\
&=x^2+\pi \left(\frac{5-2x}{\pi}\right)^2\\
&=x^2+\frac{(5-2x)^2}{\pi}
\end{align*}$$
(b) The domain of $A(x)$ is $x>0$ and $10-4x>0$ (or $x<2.5$), thus $x\in(0,2.5)$.
(c) Use a graphing utility and input $y=x^2+\frac{(5-2x)^2}{\pi}$. Then locate the lowest point and determine the value of $x$ in that point. (Refer to the image below.)
Thus, the smallest area happens when $x\approx1.40$.
