A rectangular box has a square base with an edge length of x cm and a height of h cm. The volume of the box is given by V = x2h cm3. Find the rate at which the volume of the box is changing when the edge length of the base is 12 cm, the edge length of the base is decreasing at a rate of 2 cm/min, the height of the box is 6 cm, and the height is increasing at a rate of 1 cm/min. (4 points)
Damon did this. Look at the related questions below.
i don't see it
And just in case Damon's solution does not ring your bells, I did the second one, and Steve did the third one.
Guess what, all the same way.
Steve shut up go.. either give the damn answers or go find another website to spend ur time on
To find the rate at which the volume of the box is changing, we need to use the concept of related rates.
Given:
- Edge length of the base (x) is 12 cm and decreasing at a rate of 2 cm/min.
- Height of the box (h) is 6 cm and increasing at a rate of 1 cm/min.
We are looking for the rate of change of the volume (V) with respect to time.
The volume of the box is given by V = x^2h cm^3.
To find the rate of change of the volume, we need to find dV/dt, which represents the derivative of the volume with respect to time.
To find this derivative, we can use the chain rule.
Chain rule states: dV/dt = dV/dx * dx/dt + dV/dh * dh/dt
Let's calculate each term separately:
1. dV/dx = 2xh
(Differentiating V with respect to x, treating h as a constant)
In this case, h is not changing with respect to time.
2. dx/dt = -2
(Given that the edge length of the base is decreasing at a rate of 2 cm/min)
3. dV/dh = x^2
(Differentiating V with respect to h, treating x as a constant)
In this case, x is not changing with respect to time.
4. dh/dt = 1
(Given that the height of the box is increasing at a rate of 1 cm/min)
Now, we can substitute the given values into the equation to find the rate at which the volume is changing:
dV/dt = dV/dx * dx/dt + dV/dh * dh/dt
= (2xh * -2) + (x^2 * 1)
= -4xh + x^2
Substituting x = 12 cm (given edge length) and h = 6 cm (given height):
dV/dt = -4(12)(6) + (12^2)
= -288 + 144
= -144 cm^3/min
Therefore, the rate at which the volume of the box is changing when the edge length of the base is 12 cm and the height is 6 cm is -144 cm^3/min. The negative sign indicates that the volume is decreasing.
oh, well, you really want it all, eh?
The very first one (calculus honor) has slightly different numbers, but it is exactly the same problem. Follow the solution, using your numbers.