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Complete parts a through f below to find nonnegative numbers x and y that satisfy the given requirements. Give the optimum value of P.x+y=96 and P=x^2y is maximizeda. Solve x+y=96 for y.y=b. Substitute the result from part a into the equation P=x2y for the variable that is to be maximized.P=c. Find the domain of the function P found in part b.= ? ​(Simplify your answer. Answer in interval​ notation.)d. Find dP/dx. Solve the equation dP/dx=0.dP/dx=Solve the equationx=? (Use a comma to separate answers as​ needed.)e. Evaluate P at any solutions found in part​ d, as well as the endpoints of the domain found in part c. Find​ P(0).​P(0)=? (Simplify your​ answer.)Determine ​P(64​)=? ​(Simplify your​ answer.)Find ​P(96)=? ​(Simplify your​ answer.)f. Give the maximum value of​ P, as well as the two numbers x and y for which x2y is that value.The maximum value of P is =?The values of x and y are x= and y=

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the two numbers x and y for which x2y is that value, as well as the highest possible value .P at any solutions found in part d, as well as the endpoints of the domain found as The maximum value of P is =4092, The values of x and y are x=64, y=32.

a. Solve x+y=96 for y.

y=96-x

b. In the equation P=x2y, replace the variable to be maximized with the outcome from part a.

P=x^2(96-x)

c. Find the domain of the function P found in part b.

Dom(P)=[0,96]

d. Find dP/dx. Solve the equation dP/dx=0.

dP/dx=2x(96-x)-x^2=0

Solve the equation

x=0,96

e. Evaluate P at any solutions identified in part d, as well as the endpoints of the domain found in part c.

P(0)=0

P(64)=4096

P(96)=0

f. provide the maximum value of P, as well as the two numbers x and y for which x2y is that value.

The maximum value of P is 4096

The values of x and y that maximize P are x=64, y=32

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