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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 plus y equals 81 and Pequalsx squared y is maximized a. Solve x plus y equals 81 for y. yequals 81 minus x b. Substitute the result from part a into the equation Pequalsx squared y for the variable that is to be maximized. Pequals x squared left parenthesis 81 minus x right parenthesis c. Find the domain of the function P found in part b. left bracket 0 comma 81 right bracket ​(Simplify your answer. Type your answer in interval​ notation.) d. Find StartFraction dP Over dx EndFraction . Solve the equation StartFraction dP Over dx EndFraction equals0. StartFraction dP Over dx EndFraction equals nothing

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sqdancefan

Answer:

  • y = 81-x
  • the domain of P(x) is [0, 81]
  • P is maximized at (x, y) = (54, 27)

Step-by-step explanation:

Given

  • x plus y equals 81
  • x and y are non-negative

Find

  • P equals x squared y is maximized

Solution

a. Solve x plus y equals 81 for y.

  y equals 81 minus x

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b. Substitute the result from part a into the equation P equals x squared y for the variable that is to be maximized.

  P equals x squared left parenthesis 81 minus x right parenthesis

__

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

  left bracket 0 comma 81 right bracket

__

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

  P = 81x² -x³

  dP/dx = 162x -3x² = 3x(54 -x) = 0

The zero product rule tells us the solutions to this equation are x=0 and x=54, the values of x that make the factors be zero. x=0 is an extraneous solution for this problem so ...

  P is maximized at (x, y) = (54, 27).

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