For a closed cylinder with radius r ⁢ cm and height h ⁢ cm, find the dimensions giving the minimum surface area, given that the volume is 40 cm3.

we know the surface area of the two ends of the closed cylinder is: 2 (r2) .

The surface are of the side of the cylinder is h (2r)

So the total surface area is: 2 ( r2) + h (2 r )

The volume of the cylinder is r2 h = 40 cm3

now solve this last equation for h:

so, h = 40 / (*r2)

Substitute the value for h into the expression for the total surface area:

2 ( r2) + (40 / ( r2)) (2 r )

Now Simplify the above expression we get:

2 ( r2) + (80 / r)

Now take the derivative:

4  r - 80 r -2

Set it equal to zero and solve for r:

4  r - 80 r -2 = 0

Divide by 4:

*r - 20 r -2 = 0

Multiply by r2 we get:

 r3 - 20 = 0

 r3 = 20

r3 = 20/

Hence,

r = (20 / )(1/3)

r  1.8534 cm

Substitute this value for r into h = 40 / ( r2) and evaluate h:

h = 40 / *(1.8534)2

    h  3.7067

your answer would be 10x - 1

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)