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.

$450 is the answer.

step-by-step explanation:

30% of 450 = 135

450-135=315

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.