Question

Please help me with this problem and please walk me through each step i am really confused.

Sand is falling from a rectangular box container whose base measures 40 inches by 20 inches at a constant rate of 300 cubic inches per minute.

a) how is the depth of the sand in the box changing?

b) the sand is form is forming a conical pile. At a particular moment, the pile is 23 inches high and the diameter of the base is 16 inches. The diameter of the base at this moment is increasing at 1.5 inches per minute, at this moment ,
1. how fast is the area of the circular base of the cone increasing?
2. How fast is the height of the pile increasing?

Answer 1

a) let the height of the sand in the box be h inches
V= (40)(20)(h) = 800h
dV/dt = 800 dh/dt
-300 = 800dh/dt
dh/dt = -300/800 = -3/8 inches per minute

b) part 2
V = (1/3)π r^2 h
dV/dt = (13)π r^2 dh/dt + (2/3)π r h dr/dt
plugging in our information ...
300 = (1/3)π(64)(dh/dt) + (2/3)π(8)(23)(.75)
solve for dy/dt

part 1
A = πr^2
r^2 = A/π
V = (1/3)π r^2 h
r^2 = 3V/(πh)

A/π = 3V/(πh)
A = 3V/h
dA/dt = (h(3dV/dt) - 3V dh/dt)/h^2
see where you can go from here

Answer 2

I am really confused on part 1. I don't even know what u did.
For part 2 where did u get .75 and did you mean to solve for dh/dt

Answer 3

the .75 comes from the fact that the diameter is increasing 1t 1.5in/min. The radius is half that.

and yes, he meant to solve for dh/dt, since that was the question asked.

Part 1 just uses the fact that the volume of a box is length*width*height.

can you say duh?

Answer 4

E=MC^2