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A rectangular standing-only section at the venue changes size as t increases in order 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 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

Respuesta :

Answer:[tex]3(2y-1)[/tex]

Step-by-step explanation:

Given

Length is increasing at the rate of [tex]\dot{x}=6\ ft/s[/tex]

Width is decreasing at the rate of [tex]\dot{y}=3\ ft/hr[/tex]

Area is given by

[tex]A=xy[/tex]

So rate of area is

[tex]\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}[/tex]

[tex]\frac{dA}{dt}=x\times (-3)+y\times (6)[/tex]

[tex]\frac{dA}{dt}=6y-3x[/tex]

[tex]\frac{dA}{dt}=3(2y-1)[/tex]

So, area is changing at the rate of [tex]3(2y-1)[/tex]

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