Question

A piece of wire 27 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?(b) How much wire should be used for the square in order to minimize the total area?

Answer 1

Answer:

a).length of wire for square = 15.25 m

b). length of wire for triangle = 11.75 m

Step-by-step explanation:

Length of a piece of wire = 27 m

This wire was cut into 2 pieces of length 'x' m and (27 - x) m

One was bent into an equilateral triangle and other into a square.

Area of the square shape = Side²

A = $(\frac{27-x}{4})^2$                        

Similarly, Area of the equilateral triangle = $\frac{\sqrt{3}}{4}(\text{Side})^2$

A' =$\frac{\sqrt{3}}{4}(\frac{x}{3})^{2}$

Total area of both the figures = A + A'

                                               F = $\frac{(27-x)^2}{16}+\frac{\sqrt{3}}{4}(\frac{x}{3} )^2$

Now we find the derivative of 'F' with respect to x and equate it to zero.

F' = $\frac{1}{16}(-1)[2(27-x)]+\frac{\sqrt{3}}{36}(2x)$ = 0

$\frac{\sqrt{3}}{18}(x)=\frac{1}{8}(27-x)$

x = 15.25 m

a). Length of wire required for a square = 15.25 m

b). Length of wire required for an equilateral triangle = 11.75 m

Answer 2

A) The amount of wire that should be used for the square in order to maximize the area is; 0 m

B) The amount of wire that should be used for the square in order to minimize the area is; 11.74 m

A) We are told that length of wire is 27 m.

Now, let the length cut for the square be x. Thus, length cut for the triangle will be; (27 - x) m.

Since the length for the square is x, then a side of the square is;

Length of side of square = x/4

Length of side of equilateral triangle = (27 - x)/3

Thus;

Area of square; A_s = (x/4)²

Area of equilateral triangle; A_t = ((27 - x)/3)²(¹/₄√3)

Thus, total area is;

A(x) = (x/4)² + ((27 - x)/3)²(¹/₄)√3)

⇒ A(x) = ¹/₁₆x² + ¹/₃₆((27 - x)²√3)

B) Total area is;  A(x) = ¹/₁₆x + ¹/₃₆((27 - x)²√3)

To minimize the total area, we will differentiate and equate to zero.

A'(x) = ¹/₈x - ¹/₁₈((27 - x)√3)

At A'(x) = 0, we have;

¹/₈x - ¹/₁₈((27 - x)√3) = 0

¹/₈x = ¹/₁₈((27 - x)√3)

¹/₈x = ³/₂√3 - ˣ/₁₈√3

¹/₈x + ˣ/₁₈√3 = ³/₂√3

x(¹/₈ + ¹/₁₈√3) = ³/₂√3

x = (³/₂√3)/(¹/₈ + ¹/₁₈√3)

x = 11.74 m

Now, A''(x) = ¹/₈ +  ¹/₁₈√3

This is greater than zero and so x = 11.74 m is a minimum

Thus, length cut for triangle = 27 - 11.74

length cut for triangle = 15.26 m

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