A piece of wire 10 m long is cut into two pieces. One piece is bent into a squar

Answer:
$x\approx 5.767$ and $10-x\approx 4.233$ and $x\approx 8.198$ and $10-x\approx 1.802$
Explanation:
To find the optimal wire cut, we can use mathematical reasoning. Let's denote the length of the wire used for the square as $x$ and the length used for the equilateral triangle as $10-x$.
(a) To maximize the total enclosed area, we need to maximize both the square's area and the triangle's area. The area of a square is given by $A_{\text{square}}=x^2$, and the area of an equilateral triangle is given by $A_{\text{triangle}}=\frac{\sqrt{3}}{4}(10-x{)}^2$.
To find the maximum, we can take the derivative of the total area $A_{\text{total}}=A_{\text{square}}+A_{\text{triangle}}$ with respect to $x$ and set it to zero:
$\frac{d{A}_{\text{total}}}{dx}=\frac{d(x^2)}{dx}+\frac{d(\frac{\sqrt{3}}{4}(10-x{)}^2)}{dx}=0$
Simplifying the equation, we get:
$2x-\frac{\sqrt{3}}{2}(10-x)=0$
Solving for $x$, we find:
$x=\frac{10\sqrt{3}}{4+\sqrt{3}}\approx 5.767$
Therefore, to maximize the enclosed area, the wire should be cut into two pieces with lengths approximately $x\approx 5.767$ and $10-x\approx 4.233$.
(b) To minimize the total enclosed area, we can follow the same approach. However, instead of setting the derivative to zero, we need to find the minimum value for the total area.
By taking the derivative of $A_{\text{total}}$ with respect to $x$ and solving for $x$, we find:
$x=\frac{10\sqrt{3}}{4-\sqrt{3}}\approx 8.198$
Hence, to minimize the enclosed area, the wire should be cut into two pieces with lengths approximately $x\approx 8.198$ and $10-x\approx 1.802$.

(a) To find the maximum total area enclosed, we need to maximize the sum of the areas of the square and the equilateral triangle.
The area of a square is given by the formula: $A_{\text{square}}=x^2$ square meters.
The area of an equilateral triangle is given by the formula: $A_{\text{triangle}}=\frac{\sqrt{3}}{4}{a}^2$, where $a$ is the side length of the equilateral triangle.
Since the perimeter of the equilateral triangle is $(10-x)$ meters, we can express the side length $a$ as: $a=\frac{10-x}{3}$ meters.
Substituting the value of $a$ into the formula for the area of the equilateral triangle, we have:
$A_{\text{triangle}}=\frac{\sqrt{3}}{4}{\left(\frac{10-x}{3}\right)}^2$ square meters.
The total area enclosed is the sum of the areas of the square and the equilateral triangle:
$A_{\text{total}}=A_{\text{square}}+A_{\text{triangle}}=x^2+\frac{\sqrt{3}}{4}{\left(\frac{10-x}{3}\right)}^2$ square meters.
To find the maximum total area, we need to take the derivative of $A_{\text{total}}$ with respect to $x$, set it to zero, and solve for $x$.
Taking the derivative of $A_{\text{total}}$ with respect to $x$, we get:
$\frac{d{A}_{\text{total}}}{dx}=2x-\frac{2\sqrt{3}}{36}(10-x)(-1)=2x+\frac{\sqrt{3}}{18}(10-x)$.
Setting $\frac{d{A}_{\text{total}}}{dx}$ to zero, we have:
$2x+\frac{\sqrt{3}}{18}(10-x)=0$.
Simplifying the equation, we get:
$2x+\frac{\sqrt{3}}{18}·10-\frac{\sqrt{3}}{18}x=0$.
Combining like terms, we have:
$(2-\frac{\sqrt{3}}{18})x=-\frac{\sqrt{3}}{18}·10$.
Solving for $x$, we get:
$x=\frac{-\frac{\sqrt{3}}{18}·10}{2-\frac{\sqrt{3}}{18}}$.
Simplifying the expression, we find:
$x=\frac{-10\sqrt{3}}{36-\sqrt{3}}$.
Therefore, to maximize the total area enclosed, the wire should be cut into two pieces such that $x=\frac{-10\sqrt{3}}{36-\sqrt{3}}$ meters is used for the square and $(10-x)$ meters is used for the equilateral triangle.
(b) To find the minimum total area enclosed, we can use the same expression for $A_{\text{total}}$ and follow a similar process to find the minimum. We can differentiate $A_{\text{total}}$ with respect to $x$, set it to zero, and solve for $x$. The obtained value of $x$ will give the minimum total area enclosed.