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Tentukan nilai maksimum  di daerah yang diarsir !

Tentukan nilai maksimum begin mathsize 14px style f left parenthesis x comma y right parenthesis equals 6 x plus 10 y end style di daerah yang diarsir !

 

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Jawaban:

Berdasarkan grafik, Terdapat 3 garis yang saling berpotongan 

Garis 1 melalui titik (0,6) dan (4,0)

Sehingga terbentuk persamaan garis :

begin mathsize 14px style fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction fraction numerator y minus 6 over denominator 0 minus 6 end fraction equals fraction numerator x minus 0 over denominator 4 minus 0 end fraction fraction numerator y minus 6 over denominator negative 6 end fraction equals x over 4 4 left parenthesis y minus 6 right parenthesis equals negative 6 x 4 y minus 24 equals negative 6 x 4 y equals negative 6 x plus 24 y equals negative 6 over 4 x plus 24 over 4 y equals negative 3 over 2 x plus 6 space space left parenthesis ke minus 2 space r uas space di space kali space 2 right parenthesis 2 y equals negative 3 x plus 12 bold 3 bold italic x bold plus bold 2 bold italic y bold equals bold 12 bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space bold left parenthesis bold 1 bold right parenthesis end style   

Garis 2 melalui titik (0,4) dan (4,0)

Sehingga terbentuk persamaan garis :

begin mathsize 14px style fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction fraction numerator y minus 4 over denominator 0 minus 4 end fraction equals fraction numerator x minus 0 over denominator 4 minus 0 end fraction fraction numerator y minus 4 over denominator negative 4 end fraction equals x over 4 4 left parenthesis y minus 4 right parenthesis equals negative 4 x 4 y minus 16 equals negative 4 x 4 y equals negative 4 x plus 16 y equals negative 4 over 4 x plus 16 over 4 y equals negative x plus 4 space space bold italic x bold plus bold italic y bold equals bold 4 bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold. bold space bold left parenthesis bold 2 bold right parenthesis end style   

garis 3 melalui titik (0,1) dan (-1,0)

Sehingga terbentuk persamaan garis :

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Eliminasi persamaan (1) dengan persamaan (3)

begin mathsize 14px style 3 x plus 2 y equals space 12 space open vertical bar cross times 1 close vertical bar space 3 x plus 2 y equals space 12 bottom enclose space space x minus space space y equals negative 1 space open vertical bar cross times 2 close vertical bar space 2 x minus 2 y equals negative 2 space end enclose space plus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 5 x plus 0 space equals 10 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space x equals 2 end style 

substitusi begin mathsize 14px style x equals 2 end style ke persamaan (3)

begin mathsize 14px style x minus y equals negative 1 left parenthesis 2 right parenthesis minus y equals negative 1 2 plus 1 equals y y equals 3 end style  

Terbentuk titik potong 1 :

begin mathsize 14px style rightwards double arrow left parenthesis 2 comma 3 right parenthesis end style 

Eliminasi persamaan (2) dengan persamaan (3)

begin mathsize 14px style x plus y equals space space space 4 space bottom enclose x minus y equals negative 1 space end enclose space plus space space space space space space 2 x equals 3 space space space space space space space space space space space space space space x equals 3 over 2 end style 

substitusi begin mathsize 14px style x equals 3 over 2 end style ke persamaan (3)

begin mathsize 14px style x minus y equals negative 1 open parentheses 3 over 2 close parentheses minus y equals negative 1 open parentheses 3 over 2 close parentheses plus 1 equals y y equals 3 over 2 plus 2 over 2 y equals 5 over 2 end style 

Terbentuk titik potong 2 :

begin mathsize 14px style rightwards double arrow open parentheses 3 over 2 comma 5 over 2 close parentheses end style 

Kemudian, lakukan uji titik pada daerah yang diarsir.

begin mathsize 14px style rightwards double arrow left parenthesis 4 comma 0 right parenthesis end style 

begin mathsize 14px style f left parenthesis x comma y right parenthesis equals 6 x plus 10 y f left parenthesis 4 comma 0 right parenthesis equals 6 left parenthesis 4 right parenthesis plus 10 left parenthesis 0 right parenthesis f left parenthesis 4 comma 0 right parenthesis equals 24 end style 

begin mathsize 14px style rightwards double arrow left parenthesis 2 comma 3 right parenthesis end style 

begin mathsize 14px style f left parenthesis x comma y right parenthesis equals 6 x plus 10 y f left parenthesis 2 comma 3 right parenthesis equals 6 left parenthesis 2 right parenthesis plus 10 left parenthesis 3 right parenthesis f left parenthesis 2 comma 3 right parenthesis equals 12 plus 30 f left parenthesis 2 comma 3 right parenthesis equals 42 end style 

begin mathsize 14px style rightwards double arrow open parentheses 3 over 2 comma 5 over 2 close parentheses end style 

begin mathsize 14px style f left parenthesis x comma y right parenthesis equals 6 x plus 10 y f open parentheses 3 over 2 comma 5 over 2 close parentheses equals 6 open parentheses 3 over 2 close parentheses plus 10 open parentheses 5 over 2 close parentheses f open parentheses 3 over 2 comma 5 over 2 close parentheses equals 9 plus 25 f open parentheses 3 over 2 comma 5 over 2 close parentheses equals 34 end style 

Sehingga, nilai maksimum dari begin mathsize 14px style f left parenthesis x comma y right parenthesis equals 6 x plus 10 y end style di daerah arsir adalah 42

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