minimum surface of a can

1.
Let h be the height of the can and * is a multiplication sign
Volume of cylinder = (pi) * R^2 * (h) where R = radius of the can
785= (pi)R^2 * (h)
h = 785/(pi * R^2)
Surface area of cylinder = circular areas of top and bottom + area of curved surface
A = 2pi*(R^2) + 2pi*R*h
A = 2pi*(R^2) + 2pi*R*[785/(pi * R^2)]
A = 2pi*R^2 + 1570/R

2.
Differentiate Area A with respect to R
A = 2pi*R^2 + 1570*R^(-1)
dA/dR = 4pi*R + 1570*(-1)*R^(-1-1)
dA/dR = 4pi*R – 1570R^(-2)
dA/dR = 4pi*R – 1570/R^2
Let dA/dR = 0
0 = 4pi*R – 1570/R^2
Multiply both sides by R^2
(0)*(R^2) = 4pi*R*( R^2)– 1570*R^2/R^2
0 = 4pi*R^3– 1570
4pi*R^3 = 1570
R^3 = 1570/4pi = 124.9366
R = 124.9366^(1/3) = 4.999 cm
R = 5 cm

You have to show that A is a minimum value when R = 5 cm
Take second derivative
d/dR(dA/dR) = d/dR[4pi*R – 1570R^(-2)]
d2A/dR2 = 4pi + 4710/Rr^3
Since R is positive, so d2A/dR2 is always positive, hence area is a minimum value.
Surface area is at minimum when radius of the can is 5 cm

3.
h = 785/(pi * R^2)
h = 785/(pi * 5^2)
h = 9.99 cm
h = 10 cm

4. Material used to produce the can will be at minimum since the area is minimum. Consequently, the cost of material will be at minimum. If the can is coated with a paint, so does the cost of the paint at minimum.

1. Let h be the height of can.
(pi)R^2h = 785
h = 785/piR^2
A = 2piR^2 + 2piRh
= 2piR^2 + 1570/R

2. dA/dR = 4piR - 1570/R^2
dA/dR = 0 <=> 4piR^3 - 1570 = 0
R = cuberoot(785/2pi)
dA/dR < 0 <=> R < cuberoot(785/2pi)
dA/dR > 0 <=> R > cuberoot(785/2pi)
So A attains minimum when R = cuberoot(785/2pi) cm

3. h = 785/[pi(785/2pi)^2/3] = cuberoot(3140/pi) cm

4. Cost of painting can reduce to minimum.