Answer:
The rate of change of the area of the section with respect to time = 12 ft²/hr
Step-by-step 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!!!
