Question

a plane carrying food and water to a resort island can carry a maximum of 30000 pounds and is limited in space to carrying no more than 550 cubic feet. each container of water weighs 60 pounds and takes up 1 cubic foot cargo space. each container of food weighs 175 pounds and takes up 5 cubic feet. hotels on the island will buy the food for 18 dollars a pound and the water for 4 dollars per pound. what is the optimum number of containers of each item that will maximize the revenue generated by the plane? what is the maximum revenue? let x represent the number of food containers and y represent the number of water containers. answer maximum revenue of s maximum revenue at

Explanation:

Step1: Set up the constraints

The weight constraint: \(60x + 175y\leq30000\)
The volume constraint: \(x + 5y\leq550\)
Also, \(x\geq0,y\geq0\) (non - negative number of containers)

The revenue function \(R=4\times60x+18\times175y = 240x + 3150y\)

From \(x + 5y\leq550\), we have \(x=550 - 5y\)

Substitute \(x = 550-5y\) into the weight constraint:

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Step2: Substitute \(y = 24\) into \(x=550 - 5y\)

\(x=550-5\times24=550 - 120 = 430\)

Step3: Calculate the revenue

Substitute \(x = 430\) and \(y = 24\) into the revenue function \(R=240x + 3150y\)

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Answer:

Maximum revenue of \(\$178800\) at \((x = 430,y = 24)\)