Iklan

Iklan

Pertanyaan

N i l ai mak s im u m d a r i f ( x , y ) = 300 x + 500 y y an g m e m e n u hi p er t i d ak s amaan x + 2 y ≤ 4 , x + y ≤ 3 , x ≥ 0 d an y ≥ 0 a d a l ah ...

  1. 900

  2. 1.000

  3. 1.100

  4. 1.200

  5. 1.500

Iklan

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Iklan

Pembahasan

C a r i space t i t i k space p o j o k space d a r i space p e r t i d a k s a m a a n space x plus 2 y less or equal than 4 space d a n space x plus y less or equal than 3 comma space y a i t u colon  space x plus 2 y less or equal than 4 left right arrow space left parenthesis horizontal strike 4 comma 0 right parenthesis end strike space d a n space left parenthesis 0 comma 2 right parenthesis space  x plus y less or equal than 3 left right arrow space left parenthesis 3 comma 0 right parenthesis space d a n space left parenthesis horizontal strike 0 comma 3 end strike right parenthesis  s e p e r t i space p a d a space g a m b a r

S e l a n j u t n y a space c a r i space t i t i k space p o t o n g space d a r i space k e d u a space p e r t i d a k s a m a a n space t e r s e b u t colon  bottom enclose x plus 2 y equals 4  x plus y equals 3 space space space minus end enclose  y equals 1  S u b t i t u s i k a n space n i l a i space y space k e space p e r s a m a a n space x plus y equals 3 comma space m a k a space d i p e r o l e h  x plus 1 equals 3 left right arrow x equals 2  S e h i n g g a space t i t i k space p o t o n g n y a space y a i u t space left parenthesis 2 comma 1 right parenthesis  N i l a i space m a k s i m u m space d i p e r o l e h space d e n g a n space m e n s u b s t i t u s i k a n space t i t i k minus t i t i k space t e r s e b u t space k e space f u n g s i comma space y a i t u  space f left parenthesis x comma y right parenthesis equals 300 x plus 500 y  space f left parenthesis 2 comma 0 right parenthesis equals 300 left parenthesis 2 right parenthesis plus 0 equals 600 space  f left parenthesis 0 comma 3 right parenthesis equals 0 plus 500 left parenthesis 3 right parenthesis equals 1500 space left parenthesis m a k s i m u m right parenthesis space  f left parenthesis 2 comma 1 right parenthesis equals 300 left parenthesis 2 right parenthesis plus 500 left parenthesis 1 right parenthesis equals 1100  space J a d i space n i l a i space m a k s i m u m n y a space a d a l a h space 1500

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

71

Iklan

Iklan

Pertanyaan serupa

Seorang pemilik toko sandal memiliki modal Rp4.000.000,00. Ia membeli setiap pasang sandal A Rp10.000,00 dan sandal B Rp8.000,00. Setiap pasang sandal A dan sandal B masing-masing memberi keuntungan R...

18

4.3

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia