Question

a block of ice in the shape of a cube originally has a volume of 1000 cm^3. as it melts each of the edges is dereasing at a rate of 1cm/hr. At what rate is the surface area of the cube decreasing when its volume is 27 cm^3

Answer 1

help!

Answer 2

please..

Answer 3

the surface area of a cube is 6x^2, the derivative is 12x so the rateof change when volume is 27( x= 3 ) is 12*3 = 36 cm^2 per hour

I think

Answer 4

Let x be length of edge of cube
SA = 6x^2
dSA/dx = 12x

dx/dt = -1

dSA/dt = dx/dt * dSA/dx

dSA/dt = -12x

Now Volume = x^3
x^3 = 27
x = 3

Therefore, dSA/dt = -12*3 = -36

Answer 5

thank you soo much

Answer 6

its cm^2/hr right?

Answer 7

yes

Answer 8

cool man

Answer 9

let the edge of the cube be a
da/dt=-1cm/hr
originally the volume was 1000 cm^3 so a^3=1000 a=10 cm initially surface area s=6a^2
we have to find ds/dt when v=27 cm^3 or a=3 cm
differentiate s with respect to t
ds/dt=12a da/dt ds/dt=12*3*-1
ds/dt=-36
=-36 cm^2/hr