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3 years ago

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The sum of the circumference of a circle and the perimeter of a square is 24. Find the dimensions of the circle and square that

produce a minimum total area. (Let x be the length of a side of the square and r be the radius of the circle.)

1 answer:

lyudmila [28]3 years ago

8 0

Answer:

The radius of the circle is $r=1.68\ units$

The length of the square is $x=3.36\ units$

Step-by-step explanation:

we know that

The circumference of a circle is equal to $C=2\pi r$

The perimeter of the square is equal to $P=4x$

so

$24=2\pi r+4x$

Simplify

$12=\pi r+2x$

$x=(12-\pi r)/2$ -----> equation A

The area of a circle is equal to $A=\pi r^{2}$

The area of a square is $A=x^{2}$

The total area is equal to

$At=\pi r^{2}+x^{2}$ -----> equation B

substitute equation A in equation B

$At=\pi r^{2}+[(12-\pi r)/2]^{2}$

This is a vertical parabola open upward

The vertex is the minimum

The x-coordinate of the vertex is the radius of the circle that produce a minimum area

The y-coordinate of the vertex is the minimum area

Solve by graphing

The vertex is the point (1.68, 20.164)

see the attached figure

therefore

The radius of the circle is

$r=1.68\ units$

Find the value of x

$x=(12-\pi r)/2$

assume

$\pi =3.14$

$x=(12-(3.14)*(1.68))/2$

$x=3.36\ units$

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Option A: $O(0,0), S(0,a), T(a,a), W(a,0)$

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$\begin{array}{l}{\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } S T=\sqrt{(a-0)^{2}+(a-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } T W=\sqrt{(a-a)^{2}+(0-a)^{2}}=\sqrt{a^{2}}=a} \\{\text { Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,a), T(a,a), W(a,0)$ has sides of length a.

Option B: $O(0,0), S(0,a), T(2a,2a), W(a,0)$

Now, we shall find the length of the square,

$\begin{aligned}&\text { Length } O S=\sqrt{(0-0)^{2}+(a-0)^{2}}=\sqrt{a^{2}}=a\\&\text {Length } S T=\sqrt{(2 a-0)^{2}+(2 a-a)^{2}}=\sqrt{5 a^{2}}=a \sqrt{5}\\&\text {Length } T W=\sqrt{(a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{2 a^{2}}=a \sqrt{2}\\&\text {Length } O W=\sqrt{(a-0)^{2}+(0-0)^{2}}=\sqrt{a^{2}}=a\end{aligned}$

This is not a square because the lengths are not equal.

Option C: $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$

Now, we shall find the length of the square,

$\begin{array}{l}{\text { Length OS }=\sqrt{(0-0)^{2}+(2 a-0)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } S T=\sqrt{(2 a-0)^{2}+(2 a-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } T W=\sqrt{(2 a-2 a)^{2}+(0-2 a)^{2}}=\sqrt{4 a^{2}}=2 a} \\{\text { Length } O W=\sqrt{(2 a-0)^{2}+(0-0)^{2}}=\sqrt{4 a^{2}}=2 a}\end{array}$

Thus, the square with vertices $O(0,0), S(0,2a), T(2a,2a), W(2a,0)$ has sides of length 2a.

Option D: $O(0,0), S(a,0), T(a,a), W(0,a)$

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Thus, the correct answers are option a and option d.

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