Question

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a​ minimum, what is the length of each​ part?

Answer 1

If the sum of the areas enclosed by each part is a​ minimum, the length of each​ part is $\frac{\sqrt{3} \pi b}{9+\sqrt{3} \pi}$

Explanation:

A wire is a single, usually cylindrical, flexible strand or rod of metal. There are many types of wire such as: Black : Hot wire, for switches or outlets, Red : Hot wire, for switch legs, Blue and Yellow : Hot wires, pulled in conduit, White : Always neutral, Green and Bare Copper : Only for grounding.

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle. If the sum of the areas enclosed by each part is a​ minimum, what is the length of each​ part?

An equilateral triangle is a triangle where all three sides are equal. An equilateral triangle is also equiangular that all three internal angles are also congruent to each other and are each 60°.

According to the picture attached below  

$A(x)=\pi (\frac{x}{2\pi})^2 + 1/2 \frac{b-x}{3} \frac{\sqrt{3} (b-x) }{6}$

$A(x)=\frac{x^2}{4\pi} +\frac{\sqrt{3}}{36} (b-x)^2$

$A'(x)=\frac{x}{2\pi} -\frac{\sqrt{3}}{18} (b-x) = 0$ when

$x=\frac{\sqrt{13}/18 b}{1/2\pi + \sqrt{3}/18}$

$\frac{\sqrt{3} \pi b}{9+\sqrt{3} \pi}$

Learn more about an equilateral triangle brainly.com/question/3591053

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Answer 2

1. Divide wire b in parts x and b-x. 

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A=$\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

 $2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)=$\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)=$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$=0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b-$\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$