The side length of a rectangular box with a square base is increasing at the rate of 2 ft/sec, while the height is decreasing at the rate of 2 ft/sec. At what rate is the volume of the box changing when the side length is 10 ft and the height is 12 ft? Remember to use the product rule when you find the expression for dV,dt.

Just give the answer Steve

so, did you use the product rule as advised?

v = x^2 h
dv/dt = 2xh dx/dt + x^2 dh/dt

Now just plug in your values to find dv/dt.

Note that dv/dt does not actually depend on h at all, just dh/dt.

To find the rate at which the volume of the box is changing, we need to differentiate the volume function with respect to time.

Let's assign variables:
- \(l\) : length of the side of the square base (in ft)
- \(h\) : height of the box (in ft)
- \(V\) : volume of the box (in ft³)
- \(\frac{{dl}}{{dt}}\) : rate at which the side length is changing (in ft/sec), given as \(2\) ft/sec
- \(\frac{{dh}}{{dt}}\) : rate at which the height is changing (in ft/sec), given as \(-2\) ft/sec
- \(\frac{{dV}}{{dt}}\) : rate at which the volume is changing (in ft³/sec)

The volume of the box can be expressed as \(V = l^2 \cdot h\).

Differentiating V with respect to t using the product rule:
\(\frac{{dV}}{{dt}} = 2l \cdot \frac{{dl}}{{dt}} \cdot h + l^2 \cdot \frac{{dh}}{{dt}}\)

Substituting the given values:
\(\frac{{dV}}{{dt}} = 2 \cdot 10 \cdot 2 \cdot 12 + 10^2 \cdot (-2)\)

Simplifying:
\(\frac{{dV}}{{dt}} = 40 \cdot 12 - 100 \cdot 2\)

Calculating:
\(\frac{{dV}}{{dt}} = 480 - 200\)

\(\frac{{dV}}{{dt}} = 280\) ft³/sec

Therefore, when the side length is 10 ft and the height is 12 ft, the volume of the box is changing at a rate of 280 ft³/sec.

To find the rate at which the volume of the box is changing, we can use the chain rule and the product rule.

Let's define the variables:
- s: side length of the square base of the rectangular box (ft)
- h: height of the box (ft)
- V: volume of the box (ft^3)

We are given the following information:
- ds/dt = 2 ft/sec (rate at which the side length is increasing)
- dh/dt = -2 ft/sec (rate at which the height is decreasing)
- s = 10 ft (side length at a specific moment)
- h = 12 ft (height at a specific moment)

The formula for the volume of a rectangular box is V = s^2 * h.

To find the rate at which the volume is changing (dV/dt), we need to express V in terms of t and then differentiate it with respect to t.

1. Express the volume V in terms of the side length s and height h:
V = s^2 * h.

2. Differentiate V with respect to t using the product rule:
dV/dt = (d(s^2)/dt) * h + s^2 * (dh/dt).

Applying the product rule:
dV/dt = 2s * ds/dt * h + s^2 * (dh/dt).

Substituting the given values:
s = 10 ft, h = 12 ft, ds/dt = 2 ft/sec, dh/dt = -2 ft/sec:
dV/dt = 2(10) * 2 * 12 + (10)^2 * (-2).
dV/dt = 40 * 2 * 12 + 100 * (-2).
dV/dt = 960 - 200.
dV/dt = 760 ft^3/sec.

Therefore, the volume of the box is changing at a rate of 760 ft^3/sec when the side length is 10 ft and the height is 12 ft.