Question

A piece of wire 26 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?

Answer 1

Answer:

For the maximize the total area, the all wire i.e. 26 m should be used for the square.

Step-by-step explanation:

Length of the wire = 26 m

Let Amount of wire cut for square = x

Amount of wire cut for triangle  = 26 - x

Side of the square = $\frac{x}{4}$

Area of the square = $\frac{x^{2} }{16}$  ------ (1)

Side of the triangle is given by

$a = \frac{26 - x}{3}$

Side of the triangle is $a = \frac{26 - x}{3}$

Area of the triangle is given by

$A = \frac{\sqrt{3} }{4} a^{2}$

Area of the triangle is

$A =\frac{\sqrt{3} }{36} (26 - x)^{2}$  ------- (2)

Now the total area = Area of square + Area of triangle

The total area = $\frac{x^{2} }{16} + \frac{\sqrt{3} }{36} (26 - x)^{2}$  ------- (3)

Differentiate above equation with respect to x we get

$A' = \frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x)$

Take $A' = 0$

$\frac{x}{8} - \frac{\sqrt{3} }{18} (26 - x) = 0$ ------- (4)

By solving the above equation we get

x = 11.31 m

Again take $A''$ by differentiating equation  (4)

$\frac{x}{8} + \frac{\sqrt{3} }{18} (26)$

Which is greater than zero. so the value x = 11.31 m gives the area minimum.

Thus for the maximize the total area, the all wire i.e. 26 m should be used for the square.