Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is 2picm^3

Answer and Step-by-step explanation:

Solution:

Given:

V = 2π cm3

We know that:

2π = πr2h

2π / πr2 = h

Which gives: h = 2 / r2

Minimize surface area = 2πrh + 2πr2

S = 2πrh + 2πr2

Put h = 2/ r2 in surface area, we get:

S = 2πr (2/ r2 ) + 2πr2

S = 4 π / r + 2πr2

Therefore, s(r) = 4 π / r + 2πr2

Differentiate concerning r, we have

S’(r) = -4π / r2 + 4πr

Again differentiate concerning r, we have:

S”(r) = 2 x 4 π/ r3 + 4π

For s’(r) = 0

0 = -4π / r2 + 4πr

-4π + 4πr3 = 0

r3 = 1

r = 1 cm

and s”(1) > 0 at r = 1 is point of minima.

So h = 2 π / π x1 x1

  = 2cm

Therefore minimum surface area the can should have a radius of 1cm and height of 2cm.

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510