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A large coffee urn dispenses coffee at a hospital cafeteria.The cafeteria is open from 6 am until 7 pm daily.

The rate at which coffee is added to the urn is modeled by the function e(x)=3x3+20, where the rate is measured in cups of coffee per hour since the cafeteria opened. The rate at which coffee is dispensed from the urn is modeled by the function l(x)=5x2+10, where the rate is measured in cups of coffee per hour since the cafeteria opened.

 

What does (e−l)(4) mean in this situation?

 

The rate at which the number of cups of coffee in the urn is changing 4 hours after the cafeteria opened is 302 cups per hour.

There are 302 cups of coffee in the urn 4 hours after the cafeteria opened.

----->The rate at which the number of cups of coffee in the urn is changing 4 hours after the cafeteria opened is 122 cups per hour.

There are 122 cups of coffee in the urn 4 hours after the cafeteria opened.

 
 Jan 10, 2019
 #1
avatar+94526 
+2

Rate that coffee is being added to the urn  after 4 hours is  3(4)^3 + 20   = 212 cups/hr

 

Rate that coffee is being dispensed  from the urn after 4 hours = 5(4)^2 + 10  =   90 cups/hr

 

So

 

(e - l)(4)   =   212 - 90   =  122  [ this means that  after 4 hours, coffee is being added to the urn at 122 cups / hr ]

 

 

cool cool cool

 
 Jan 10, 2019

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