Min p = 10x + 26y
subject to
x+y< 6
5x+y> 10
x+5y> 14

Step-by-step explanation:

•x + y < 6

(6,6)

• 5x + y = 10

(2,10)

•x + 5y = 14

(14,14/5)

intersection I

x + 5y = 14

5x + y = 10

____________ •2

5x + 25y = 70

5x + y = 10

____________ _

24y = 60

y = 2 1/2 = 2,5

substitusion

5x + y = 10

5x + 2,5 = 10

5x = 7,5

x = 1,5

(1,5 , 2,5)

Intersection II

x + y = 6

5x + y = 10

_____________ _

-4x = -4

x = 1

x + y = 6

y = 6 - 1

y = 5

(1 , 5)

Intersection III

x + y = 6

x + 5y = 14

___________ _

-4y = -8

y = 2

x + y = 6

x = 6 - 2

x = 4

(4 , 2)

p = 10x + 26y

I.(1,5 , 2,5) = 10(1,5) + 26(2,5) = 15 + 65 = 80. (MINIMUM)

II.(1,5) = 10(1) + 26(5) = 10 + 130 = 140(MAXIMUM)

III.(4,2) = 10(4) + 26(2) = 40 + 52 = 92

SO,MINIMUM VALUE ------> 80

the answer is y = 11.54

hope this : )