A coffee merchant sells three blends of coffee. A bag of the house blend contains 300 grams of Colombian beans and 300 grams of French roast beans. A bag of the special blend contains 200 grams of Colombian beans, 200 grams of Kenyan beans, and 100 grams of French roast beans. A bag of the gourmet blend contains 300 grams of Colombian beans, 200 grams of Kenyan beans, and 300 grams of French roast beans. The merchant has on hand 39 kilograms of Colombian beans, 15 kilograms of Kenyan beans, and 36 kilograms of French roast beans. If he wishes to use up all of the beans, how many bags of each type of blend can be made

The Merchant can Use All of His Beans Making 65 Bags Of House Blend , 30 Bags Of special Blend And 45 Bags Of Gourmet Blend .

Step-by-step explanation:

Abbreviation Used ( Colombian Bean = C , Kenyan Bean = K , French Roast Beans = F , House Blend = H , Special Blend = S , Gourmet Blend = G)

Now According to Question, We have 39kg of C , 15kg of K , 36kg of F

300gm in H + 200gm in S + 300gm in G = Total Colombian Beans(39000gm) ↔ Eq. 1

200gm in G + 200gm in S  = Total Kenyan Bean(15000gm) ↔ Eq. 2

300gm in H + 100gm in S + 300 in of G = Total FrenchRoast Beans(36000gm) ↔ Eq. 3

if we add all three Equations we get

600gm in H  + 500gm in S + 800gm in G = C+K+F(90000gm)↔Eq. 4

Now Subtract Eq. 1 - Eq. 3 We Get

100*S=3000

Total bag of Special Blend is 30

Now Put the value of S=30 in above equations(Eq. 1 & Eq. 4)

we get

3H+3G=330↔(eq. 5)  &   6H+8G=750↔(eq. 6)

solve these equation by subtracting (eq. 6 - 2*eq.5)

we get 2G=90  ⇔ Total bag of Gourmet Blend is 45

now By Putting G value in Eq. 6

We Get 6H=390 ⇔ Total bag of House Blend is 65

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

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