Answer :
1) Find the perimeters of the square and the triangle.
Suppose you divide the wire into 2 pieces that measure x and y. x is used to make the square and y is used to make the triangle, according to the figure:
x is the perimeter of the square;
y is the perimeter of the triangle.
Since the total wire is 30 inches long:
x + y = 30 (equation 1).
2) Find the area of the geometric figures.
Square:
Since a square has 4 equal sides, each side will measure x/4 (4*x/4=x)
And, since the area (As) of a square whose side measures a is:
[tex]As=a^2[/tex]The area of the square is:
[tex]As=(\frac{x}{4})^2=\frac{x^2}{16}[/tex]Equilateral Triangle:
Since an equilateral triangle has 3 equal sides, each side will measure y/3 (3*y/3 = y).
And, since the area (At) of an equilateral triangle whose side measures b is:
[tex]At=\frac{\sqrt[]{3}}{4}\cdot b^2[/tex]The area of the triangle is:
[tex]\begin{gathered} At=\frac{\sqrt[]{3}}{4}\cdot(\frac{y}{3})^2 \\ At=\frac{\sqrt[]{3}}{4}\cdot\frac{y^2}{9}=\frac{\sqrt[]{3}}{36}\cdot y^2 \end{gathered}[/tex]3) Find the total area.
The total area (A) is the sum of the areas:
[tex]\begin{gathered} A=As+At \\ A=\frac{x^2}{16}+\frac{\sqrt[]{3}}{36}\cdot y^2 \end{gathered}[/tex]4) Isolate y in equation 1 and substitute in the equation for the area.
x + y = 30
y = 30 - x
[tex]A=\frac{x^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-x)^2[/tex]5) The maximum or minimum value of A happens at x when dA/dx = 0.
So, derivate A:
[tex]\begin{gathered} \frac{dA}{dx}=\frac{2x}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-x)\cdot2\cdot(-1) \\ \frac{dA}{dx}=\frac{x}{8}-\frac{\sqrt[]{3}}{18}(30-x) \end{gathered}[/tex]And do dA/dx = 0 to find x.
[tex]0=\frac{x}{8}-\frac{\sqrt[]{3}}{18}(30-x)[/tex]Then, isolate x:
[tex]\begin{gathered} 0=\frac{x}{8}-\frac{\sqrt[]{3}}{18}\cdot30+\frac{\sqrt[]{3}}{18}\cdot x \\ \frac{\sqrt[]{3}}{18}\cdot30=\frac{x}{8}+\frac{\sqrt[]{3}}{18}\cdot x \\ 2.89=0.3x \\ x=\frac{2.89}{0.22} \\ x=13 \end{gathered}[/tex]6) Evaluate the equation for x at the exremes x = 0; y = 30, and at x = 13.
For x = 0
[tex]\begin{gathered} A=\frac{x^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-x)^2 \\ \frac{0^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-0)^2 \\ =\frac{\sqrt[]{3}}{36}\cdot900 \\ =43.3 \end{gathered}[/tex]For x = 30
[tex]\begin{gathered} A=\frac{x^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-x)^2 \\ A=\frac{30^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-30)^2 \\ =\frac{900}{16}+\frac{\sqrt[]{3}}{36}\cdot0 \\ =56.25 \end{gathered}[/tex]And for x =13
[tex]\begin{gathered} A=\frac{13^2}{16}+\frac{\sqrt[]{3}}{36}\cdot(30-13)^2 \\ A=\frac{169}{16}+\frac{\sqrt[]{3}}{36}\cdot17^2 \\ A=\frac{169}{16}+\frac{\sqrt[]{3}}{36}\cdot289 \\ A=24.47 \end{gathered}[/tex]7) Compare the results:
As you can see in step 6, when x = 13, the area is minimized, while the area is maximized when x = 30.
The greatest area will be reached when x = 30. That means y = 0 (x + y =30).
Answer:
The greatest area is reached when all the wire is used to make the square.
To make the 2 figures, we wire should be cut as nearest as possible to 30, and the greatest piece should be used to make the square.
