The rate at which raw sewage enters a treatment tankis given by E(t)= 850 + 715cos((pi x^2/(9)) gallons per hour for 0 -

Question

The rate at which raw sewage enters a treatment tankis given by E(t)= 850 + 715cos((pi x^2/(9)) gallons per hour for 0 -< t -< 4 hours. Treated sewage is removed from the tank at the constant rate of 645gallons per hour. The treatment tank is empty at time t = 0.

For 0 -< t _< 4, at what time t is the amount of sewage in the treatment tank greatest? To the nearest gallon, what is the maximum amount of sewage in the tank? Justify your answers.

arabpride · Answer

You posted this question in history section, when it should be posted in math section. Please repost your question there, and close this one. 
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