Question

A piece of wire 6 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire should be used for the square in order to maximize the total area? Correct: Your answer is correct. m (b) How much wire should be used for the square in order to minimize the total area? Incorrect: Your answer is incorrect. m

Answer 1

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-

$x+y=6$ or $x= 6-y$

The length of the side of square is $\frac{x}{4}$

The length of the side of triangle is $\frac{y}{3}$

We have to find the area.

Area of square = $side^{2}$ = $(\frac{x}{4})^{2}$ = $\frac{x^{2}}{16}$

Area of triangle = $a^{2} \frac{\sqrt{3} }{4}$

= $(\frac{y}{3} )^{2} \times\frac{\sqrt{3} }{4}$

= $\frac{y^{2} }{9}\times \frac{\sqrt{3} }{4}$

Now differentiating the function to maximize the total area-

A = $\frac{x^{2}}{16}+(\frac{y^{2} }{9} \times \frac{\sqrt{3} }{4})$

Substituting x=6-y

$\frac{(6-y)^{2}}{16}+(\frac{y^{2} }{9} \times \frac{\sqrt{3} }{4})$

Differentiating the function with respect to y, we get

A(y) = $\frac{-(6-y)}{8}+\frac{y}{9} \times \frac{\sqrt{3} }{2}$

Now equating y to 0, we get

y = $\frac{54}{9+4\sqrt{3} }$

The function reaches the minimum at y = $\frac{54}{9+4\sqrt{3} }$

We can find the maximum area at x=0 and x=6

$A=\frac{6^{2}}{16}+\frac{0^{2}}{9}\times\frac{\sqrt{3}}{4}$ = $\frac{9}{4}$

$A=\frac{0^{2}}{16}+\frac{6^{2}}{9}\times\frac{\sqrt{3}}{4}$ = $\sqrt{3}$

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 = $6-\frac{54}{9+4\sqrt{3} }$

= $\frac{24\sqrt{3} }{9+4\sqrt{3} }$

Therefore, the lengths of x and y can be used to minimize the total area.

x=$\frac{24\sqrt{3} }{9+4\sqrt{3} }$

y=$\frac{54}{9+4\sqrt{3} }$