Question

The volume of a rectangular box with a square base remains constant at 1000cm^3 as the area of the base increases at a rate of 2 cm^2/sec. Find the rate at which the height of the box is decreasing when each side of the base is 19cm long. (do not round the answer)

Answer 1

V = base area * height.

1000 = A * h

Take the derivative of both sides with respect to t using the chain rule and product rule:
\[0 = \frac{dA}{dt}(h) +\frac{dh}{dt}(A)\]

Answer 2

We are given that
dA/dt = 2
Side is 19 so base area = 19^2 = 361

Answer 3

one more ting.
Vol = 1000
A*h = 1000
361*h = 1000
find h.
Gives you height.

insert the values in the product rule.

Answer 4

so the height would be decreasing at a rate of 2.77cm^2/sec?

Answer 5

well it looks like one of my choices are
\[A. \frac{ 2000 }{ 6859 }cm/\sec\]
\[B. \frac{ 2 }{ 361 }cm/\sec\]
\[C.\frac{ 2000 }{ 130321 }cm/\sec\]
\[D. \frac{ 1000 }{ 361 }cm/\sec\]
So would it be D then???

Answer 6

No!! That just gives you the height.
Insert the values in the product rule:

Answer 7

so if I take
\[361*\frac{ d }{ dt }\frac{ 1000 }{ 361 }+\frac{ 1000 }{ 361 }\frac{ d }{ dt }(361)=\frac{ d }{ dt }(1000)\]

Answer 8

I still don't understand