homeschool0123

25.06.2019 • 

Mathematics

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

Ответ:

Miles1111

10.03.2022

Mathematics

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

Step-by-step explanation:

Given

x plus y equals 81x 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

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

  left bracket 0 comma 81 right bracket

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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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Ответ:

20068513

05.04.2021

Mathematics

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2 5/6 = 17.6 - hope it help

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