Question

A rectangular "standing-only" section at the venue changes size as increases in order to manage the flow of people. Let represent the length, in feet, of the section, and let represent the width, in feet, of the section. The length of the section is increasing at a rate of 6 feet per hour, and the width of the section is decreasing at a rate of 3 feet per hour. What is the rate of change of the area of the section with respect to time when and ? Indicate units of measure.

Answer 1

The rate of change of the area of the section with respect to time = 12 ft²/hr

Explanation:

Complete Question

A rectangular "standing only" section at the venue changes size as t increases in onder to manage the flow of people, let x represent the length, in feet, of the section, and let y represent the width, in feet of the section. The length of the section is increasing at a rate of 6 feet/hour and the width of the section is decreasing at a rate of 3 feet/hour. What is the rate of change of the area of the section with respect to time when x=16 & y=10?

The area of a rectangular section is given as

Area = Length × Width

Area = A

Length = x

Width = y

A = xy

We now nred the time Rate of change of the area. So, we take the time derivative of both sides.

A = xy

(d/dt) (A) = (d/dt) (xy)

Using product rule for the Right hand side

(dA/dt) = x(dy/dt) + y(dx/dt)

- The length of the section is increasing at a rate of 6 feet/hour, that is, (dx/dt) = 6 ft/hr

- The width of the section is decreasing at a rate of 3 feet/hour, that is, (dy/dt) = -3 ft/hr (minus sign because it is decreasing)

x = 16 ft

y = 10 ft

(dA/dt) = x(dy/dt) + y(dx/dt)

(dA/dt) = (16)(-3) + (10)(6) = -48 + 60 = 12 ft²/hr

Hope this Helps!!!