the volume of a rectangular box with a square base remains constant at 500 cm^3 as the area of the box increases at a rate of 3cm^2/sec. find the rate at which the height of the box is decreasing when each side of the base is 18 cm long.

To find the rate at which the height of the box is decreasing, we need to use related rates.

Let's denote the length of the side of the square base as s, and let h represent the height of the box.

We know that the volume of the rectangular box is constant at 500 cm^3, so we can write the equation:

V = s^2 * h = 500

Now, we are given that the area of the box is increasing at a rate of 3 cm^2/sec. This means:

dA/dt = 3 cm^2/sec

The area of the box can be expressed as the sum of the areas of the base and the four sides:

A = s^2 + 4sh

Since we want to find dh/dt, we need to differentiate the equation with respect to t:

dA/dt = 2s(ds/dt) + 4h(dh/dt)

Now, we can substitute the known values:

3 = 2(18)(ds/dt) + 4(18)(dh/dt)

Simplifying further:

3 = 36(ds/dt) + 72(dh/dt)

Now, we can rearrange the equation to isolate dh/dt:

72(dh/dt) = 3 - 36(ds/dt)

dh/dt = (3 - 36(ds/dt))/72

We know that ds/dt represents the rate at which the sides of the base are increasing, but in this case, we are given that each side of the base is 18 cm long. Therefore, ds/dt = 0.

Substituting that value back into the equation:

dh/dt = (3 - 36(0))/72
dh/dt = 3/72
dh/dt = 1/24 cm/sec

Therefore, the rate at which the height of the box is decreasing when each side of the base is 18 cm long is 1/24 cm/sec.

To find the rate at which the height of the box is decreasing, we can use the concept of related rates. Let's set up the problem step by step:

1. Start by determining the dimensions of the box. We know that the volume remains constant at 500 cm^3, so we have V = 500 cm^3.

2. Since the base of the box is square, let's denote the side length of the base as 's' and the height of the box as 'h'.

3. Based on the given information, we have (side length of the base)^2 * height = volume. Substituting the known values, we get s^2 * h = 500.

4. Now, let's calculate the rate at which the volume is changing with respect to time, which is dV/dt. Since the volume remains constant at 500 cm^3, the rate of change of volume is 0.

5. Taking the derivative of the equation s^2 * h = 500 with respect to time, we can get the related rates equation: 2s * ds/dt * h + s^2 * dh/dt = 0.

6. Since we are given that the area of the box is increasing at a rate of 3 cm^2/sec, we know ds/dt = 3 cm^2/sec.

7. We are also given that each side of the base has a length of 18 cm. Therefore, s = 18 cm.

8. Plugging in the known values, we have 2 * 18 * 3 * h + 18^2 * dh/dt = 0.

9. Simplifying the equation, we get 108h + 324 * dh/dt = 0.

10. Finally, solving for dh/dt (the rate at which the height is decreasing), we get dh/dt = -108h / 324.

11. Now let's substitute the value of h when each side of the base is 18 cm. Since the box is rectangular, the height will also be equal to 18 cm.

12. Substituting h = 18 cm in the equation, we have dh/dt = -108 * 18 / 324.

13. Calculating the answer, we get dh/dt = -6 cm/sec.

Therefore, the rate at which the height of the box is decreasing is -6 cm/sec when each side of the base is 18 cm long.

500=areabase* h

take the derivative.

0=d areabase/dt * h+ areabase*dh/dt

solve for dh/dt when
areabase=18^2
h= 500/18^2
dareabase/dt=3 cm^2/sec