let the radius of circle be r
let the side of the square be x
4x + 2πr = 10
2x + πr = 5
x = (5 - πr)/2
total area = πr^2 + x^2
replace the x with (5 - πr)/2
careful with the squaring.
a) Differentiate, set that equal to zero and solve for r
sub back into the expression for x to find the side of the square.
b) for a max, since a circle is the largest shape possible for a given perimeter, use all of the wire for the circle.
Hi pls help me i dunno what to do show me complete solution please please Thank you.
a piece of wire 10ft. long is cut into two pieces , one piece is bent into the shape of a circle and the other into the shape of a square. How should the wire be cut so that:
a.)The combined area of the two figures is as small as possible?
b.)The combined area of the two figures is as large as possible?
Got Answer from my notes but no solution:
a.) Radius of circle = 5/(pi+4);
Length of side of square = 10/(pi+4)
b.) Radius of circle = 5/pi and there is no square
2 answers
radius of circle = r
side of square = s
10 = 4 s + 2 pi r
so s = 2.5 - .5 pi r
A = pi r^2 + s^2
or
A = pi r^2 + 6.25 - 2.5 pi r + .25 pi^2 r^2
A = (.25 pi +1)pi r^2 -2.5 pi r + 6.25
I am going to use calculus, you can complete the square to find vertex of parabola
dA/dr = 0 for max or min in domain so
0 = 2(pi)(1+.25 pi)r -2.5 pi
0 = (1+.25 pi)r - 1.25
r = 1.25/(1 + .25 pi)
or
r = 5/(4 + pi) agreed
side of square = s
10 = 4 s + 2 pi r
so s = 2.5 - .5 pi r
A = pi r^2 + s^2
or
A = pi r^2 + 6.25 - 2.5 pi r + .25 pi^2 r^2
A = (.25 pi +1)pi r^2 -2.5 pi r + 6.25
I am going to use calculus, you can complete the square to find vertex of parabola
dA/dr = 0 for max or min in domain so
0 = 2(pi)(1+.25 pi)r -2.5 pi
0 = (1+.25 pi)r - 1.25
r = 1.25/(1 + .25 pi)
or
r = 5/(4 + pi) agreed