width of rectangle ---- x
length of rectangle --- 3x
radius of circle ----- r
so 2x + 6x + 2πr = 16
4x + πr = 8 ----> x = 2 - πr/4
let S be the sum of the areas of the rectangle , (you had "square" , I assume that was an error)
S = πr^2 + 3x^2
= πr^2 + 3(1 - πr/4)^2
= πr^2 + 3 - 3πr/2 + π^2 r^2 / 16
dS/dr = 2πr - 3π/2 + (1/8)π^2 r
= 0 for a max/min
2πr - 3π/2 + (1/8)π^2 r
16πr - 12π + π^2 r = 0
divide by π
16r - 12 + πr = 0
r(16+π) = 12
r = 12/(16+π)
then x = 2 - πr/4 = .....
ratio of longer side : radius
= 3x / r
= 3x(16+π)/12
= x(16+π) : 4
check my arithmetic
b)
"add ... to the value that makes the area minimum" ???
do we add it to the x or to the r ??
You decide, find the new dimensions, then calculate the current and the new area.
That would be just arithmetic.
Take a look at the many question of the same type below in Related Questions.
A piece of wire 16cm long is cut into 2 lengths, one of which is bent into a circle, the other into a rectangle with one side three times the other.
a) Determine the ratio of the longer length of the rectangle to the radius of the circle if the sum of the areas of the circle and the square is minimum and find this minimum area.
b) If you add 6pi/(3pi+16) to the value that makes the area minimum, how much area will increase?
1 answer
1 answer