Respuesta :
Answer:
Let x be the perimeter of square.
Let y be the perimeter of equilateral triangle.
As both the shapes are made from a single wire, we can say that-
[tex]x+y=6[/tex] or [tex]x= 6-y[/tex]
The length of the side of square is [tex]\frac{x}{4}[/tex]
The length of the side of triangle is [tex]\frac{y}{3}[/tex]
We have to find the area.
Area of square = [tex]side^{2}[/tex] = [tex](\frac{x}{4})^{2}[/tex] = [tex]\frac{x^{2}}{16}[/tex]
Area of triangle = [tex]a^{2} \frac{\sqrt{3} }{4}[/tex]
= [tex](\frac{y}{3} )^{2} \times\frac{\sqrt{3} }{4}[/tex]
= [tex]\frac{y^{2} }{9}\times \frac{\sqrt{3} }{4}[/tex]
Now differentiating the function to maximize the total area-
A = [tex]\frac{x^{2}}{16}+(\frac{y^{2} }{9} \times \frac{\sqrt{3} }{4})[/tex]
Substituting x=6-y
[tex]\frac{(6-y)^{2}}{16}+(\frac{y^{2} }{9} \times \frac{\sqrt{3} }{4})[/tex]
Differentiating the function with respect to y, we get
A(y) = [tex]\frac{-(6-y)}{8}+\frac{y}{9} \times \frac{\sqrt{3} }{2}[/tex]
Now equating y to 0, we get
y = [tex]\frac{54}{9+4\sqrt{3} }[/tex]
The function reaches the minimum at y = [tex]\frac{54}{9+4\sqrt{3} }[/tex]
We can find the maximum area at x=0 and x=6
[tex]A=\frac{6^{2}}{16}+\frac{0^{2}}{9}\times\frac{\sqrt{3}}{4}[/tex] = [tex]\frac{9}{4}[/tex]
[tex]A=\frac{0^{2}}{16}+\frac{6^{2}}{9}\times\frac{\sqrt{3}}{4}[/tex] = [tex]\sqrt{3}[/tex]
Therefore, you should use x = 0m or x = 6m for square to get the total area to be maximum.
Now we can evaluate for x
We know x = 6-y
x = [tex]6-\frac{54}{9+4\sqrt{3} }[/tex]
= [tex]\frac{24\sqrt{3} }{9+4\sqrt{3} }[/tex]
Therefore, the lengths of x and y can be used to minimize the total area.
x=[tex]\frac{24\sqrt{3} }{9+4\sqrt{3} }[/tex]
y=[tex]\frac{54}{9+4\sqrt{3} }[/tex]
