The dimensions of the circle and square that produce a minimum total area are 1.12 and 2.24 respectively
Represent the side length of the square with s, and the radius of the circle with r.
So, the perimeter (P) and the area (A) of the shape are:
[tex]P =4s + 2\pi r[/tex]
[tex]A =s^2 + \pi r^2[/tex]
The perimeter is 16. So, we have:
[tex]4s + 2\pi r = 16[/tex]
Divide through by 4
[tex]s + 0.5\pi r = 4[/tex]
Make s the subject
[tex]s = 4 - 0.5\pi r[/tex]
Substitute [tex]s = 4 - 0.5\pi r[/tex] in [tex]A =s^2 + \pi r^2[/tex]
[tex]A = (4 - 0.5\pi r )^2 + \pi r^2[/tex]
Expand
[tex]A = 16 - 4\pi r + 0.25(\pi r)^2 + \pi r^2[/tex]
This gives
[tex]A = 16 - 4\pi r + (0.25\pi^2 + \pi )r^2[/tex]
Differentiate with respect to r
[tex]A' = 0 - 4\pi + 2(0.25\pi^2 + \pi )r[/tex]
[tex]A' = - 4\pi + 2(0.25\pi^2 + \pi )r[/tex]
Set to 0
[tex]- 4\pi + 2(0.25\pi^2 + \pi )r =0[/tex]
Add 4pi to both sides
[tex]2(0.25\pi^2 + \pi )r =4\pi[/tex]
Divide both sides by 2
[tex](0.25\pi^2 + \pi )r =2\pi[/tex]
Make r the subject
[tex]r =\frac{2\pi}{(0.25\pi^2 + \pi )}\\[/tex]
Factor out pi
[tex]r =\frac{2\pi}{\pi(0.25\pi + 1 )}[/tex]
Cancel out the common factors
[tex]r =\frac{2}{0.25\pi + 1 }[/tex]
Express pi as 3.14
[tex]r =\frac{2}{0.25\times 3.14 + 1 }[/tex]
[tex]r =\frac{2}{1.785}[/tex]
Divide
[tex]r =1.12[/tex]
Recall that:
[tex]s = 4 - 0.5\pi r[/tex]
This gives
[tex]s =4 -0.5 \times \pi \times 1.12[/tex]
This gives
[tex]s =4 -0.5 \times 3.14 \times 1.12[/tex]
[tex]s =2.24[/tex]
Hence, the dimensions of the circle and square that produce a minimum total area are 1.12 and 2.24 respectively
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