On a certain workday, the rate, in tons per hour, at which unprocessed gravel ar

Question

On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by G(t) = 90 + 45cos (t^2/18), where t is measured in hours and 0 lessthanorequalto t lessthanorequalto 8. At the beginning of the workday (t = 0), the plant has 500 tons of unprocessed gravel. During the hours of operation, 0 lessthanorequalto t lessthanorequalto 8, the plant processes gravel at a constant rate of 100 tons per hour. What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer.

Explanation / Answer

amount of unprocessed gravel at plant during hours of operation ,A(t)=[0 to t](G(t) -100)dt

A(t)=[0 to t]((90+45cos(t2/18) -100)dt

A(0)=500

at relative maximum A'(t)=0,A"(t)<0

A'(t)=((90+45cos(t2/18) -100),A"(t)=0+45(-sin(t2/18))(2t/18)=-5tsin(t2/18)

A'(t)=0

(90+45cos(t2/18) -100=0

45cos(t2/18) =10

cos(t2/18) =10/45

(t2/18) =cos-1(10/45)

t=[18cos-1(10/45)]

t=4.92348

A"(4.92348)=-5*4.92348sin(4.923482/18)=-24<0

A(0)=500

A(4.92348)=500+[0 to 4.92348]((90+45cos(t2/18) -100)dt=635.376

A(8)=500+[0 to 8]((90+45cos(t2/18) -100)dt=525.551

maximum amount of unprocessed gravel =635.376 tons

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