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Precalculus
Michael Sullivan
Edition 4
Chapter 2, Section 2.6
A wire 10 meters long is to be cut into 2 pieces. One piece will be shaped as an equilateral triangle, and the other piece will be shaped as a circle.
a) Express the total area A enclosed by the pieces of wire as a function of the length x of a side of the equilateral triangle.
b) What is the domain of A?
c) Graph A=A(x); for what value is A the smallest?
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2.6 Question 12
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Re: 2.6 Question 12
by nycmath » Sat Sep 19, 2026 7:23 pm
Part A
TRIANGLE:
Let x = side length of equilateral triangle.
Let 10 meters = the total length of the wire.
We want this:
A(x) = area of equilateral triangle + area of circle
The perimeter of the triangle = 3(side) or 3x.
The area of an equilateral triangle is given by
A = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex]
CIRCLE
The total length of the wire is given to be 10 meters.
The problem states that 10 meters is divided into two parts.
10 meters = 10 and 10 - 3x.
The circumference of a circle is found using C = 2[tex]\pi[/tex]r.
We need to solve for r.
Set C = 10 - 3x
2[tex]\pi[/tex]r = 10 - 3x
Solving for r, I get this:
r = (10 - 3x)/2[tex]\pi[/tex]
The area of a circle is found by using
A = [tex]\pi r^{2 }[/tex]
Plugging the value of r into the area of a circle formula we get A of circle = [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
We want A(x) = area of equilateral triangle + area if circle.
A(x) = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex] + [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
You say?
TRIANGLE:
Let x = side length of equilateral triangle.
Let 10 meters = the total length of the wire.
We want this:
A(x) = area of equilateral triangle + area of circle
The perimeter of the triangle = 3(side) or 3x.
The area of an equilateral triangle is given by
A = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex]
CIRCLE
The total length of the wire is given to be 10 meters.
The problem states that 10 meters is divided into two parts.
10 meters = 10 and 10 - 3x.
The circumference of a circle is found using C = 2[tex]\pi[/tex]r.
We need to solve for r.
Set C = 10 - 3x
2[tex]\pi[/tex]r = 10 - 3x
Solving for r, I get this:
r = (10 - 3x)/2[tex]\pi[/tex]
The area of a circle is found by using
A = [tex]\pi r^{2 }[/tex]
Plugging the value of r into the area of a circle formula we get A of circle = [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
We want A(x) = area of equilateral triangle + area if circle.
A(x) = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex] + [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
You say?
Re: 2.6 Question 12
by nycmath » Sat Sep 19, 2026 7:32 pm
Part B
Let D = Domain of A.
The side length x must be [tex]\ge[/tex]0. Also, the length of the wire used for the perimeter of the equilateral triangle (our 3x) CANNOT be greater than the total length of the wire which is given to be 10 meters.
So, 3x [tex]\le \frac{10}{3}[/tex].
Isolating x, we get:
x [tex]\le \frac{10}{3}[/tex]
D = [0, 10/3]
You say?
Let D = Domain of A.
The side length x must be [tex]\ge[/tex]0. Also, the length of the wire used for the perimeter of the equilateral triangle (our 3x) CANNOT be greater than the total length of the wire which is given to be 10 meters.
So, 3x [tex]\le \frac{10}{3}[/tex].
Isolating x, we get:
x [tex]\le \frac{10}{3}[/tex]
D = [0, 10/3]
You say?
Re: 2.6 Question 12
by nycmath » Sat Sep 19, 2026 7:39 pm
Part C
Graph:
A(x) = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex] + [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
The smallest value for x is about 2.07736.
You say?
Graph:
A(x) = [tex]\frac{ \sqrt{3} }{4} x^{2 }[/tex] + [tex]\frac{(10-3x)^2}{4 \pi }[/tex]
The smallest value for x is about 2.07736.
You say?
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