a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie…

Explanation:

Step1: Define Variables

Let ( x ) be the number of tropical smoothies and ( y ) be the number of sport smoothies.

Step2: Set Up Constraints

• Orange juice constraint: ( 9x + 3y \leq 210 ) (simplify: ( 3x + y \leq 70 ))
• Milk constraint: ( 6x + 3y \leq 150 ) (simplify: ( 2x + y \leq 50 ))
• Non - negative constraints: ( x\geq0,y\geq0 )

Step3: Objective Function

The revenue function ( R = 4x + 3y ) (we want to maximize ( R ))

Step4: Find Intersection of Constraints

Subtract the second simplified constraint from the first: ((3x + y)-(2x + y)=70 - 50) (3x + y-2x - y = 20) (x = 20) Substitute ( x = 20 ) into ( 2x + y=50 ): (2\times20+y = 50) (40 + y=50) (y = 10)

Step5: Evaluate Objective Function at Vertex

Vertices of the feasible region are ((0,0)), ((0,50)), ((20,10)), ((\frac{70}{3},0)\approx(23.33,0))

• At ((0,0)): ( R=4\times0 + 3\times0=0)
• At ((0,50)): ( R=4\times0+3\times50 = 150) (but check orange juice: (9\times0 + 3\times50=150\leq210), but milk is (6\times0+3\times50 = 150) which is within limit, but wait, when ( y = 50), from orange juice constraint ( 3x+50\leq70\Rightarrow x\leq\frac{20}{3}\approx6.67), so our earlier intersection was wrong. Wait, let's re - solve the system of inequalities. Solve ( 3x + y=70) and ( 2x + y=50) Subtract the second equation from the first: (x = 20), then ( y=70 - 3\times20=10). The vertex ((0,50)): check orange juice (9\times0+3\times50 = 150\leq210), milk (6\times0 + 3\times50=150) (which is equal to the milk container). But when ( x = 0,y = 50), revenue is (3\times50=150). When ( x = 20,y = 10): ( R=4\times20+3\times10=80 + 30=110). When ( x=\frac{70}{3}\approx23.33,y = 0): ( R=4\times\frac{70}{3}+3\times0=\frac{280}{3}\approx93.33) Wait, I made a mistake earlier. Let's re - check the milk constraint. The milk constraint is (6x + 3y\leq150), divide by 3: (2x + y\leq50). The orange juice constraint: (9x+3y\leq210), divide by 3: (3x + y\leq70) The feasible region is bounded by (x\geq0,y\geq0), (2x + y\leq50) and (3x + y\leq70) The intersection of (2x + y = 50) and (3x + y=70) is at (x = 20,y = 10) The intersection of (2x + y=50) and (x = 0) is (y = 50) The intersection of (3x + y=70) and (y = 0) is (x=\frac{70}{3}\approx23.33) The intersection of (x = 0) and (y = 0) is ((0,0)) Now, check the orange juice when ( y = 50): (9x+3\times50\leq210\Rightarrow9x\leq60\Rightarrow x\leq\frac{20}{3}\approx6.67) So the vertex ((0,50)) is not in the feasible region defined by both constraints. The correct vertices are:

1. ((0,0)): ( R = 0)
2. ((0,50)): But (x) must be non - negative, and when (y = 50), (x) can be at most (\frac{20}{3}), so this is not a vertex of the feasible region (since the feasible region is the set of points that satisfy both constraints). The correct vertices are found by the intersection of the two main constraints and the axes. The two main constraints are (3x + y=70) and (2x + y=50) (intersection at ((20,10))), (2x + y=50) and (x = 0) (intersection at ((0,50)), but as we saw, this point does not satisfy the orange juice constraint when (x) is non - negative? Wait no, the orange juice constraint is (9x+3y\leq210), when (x = 0,y = 50), (9\times0+3\times50 = 150\leq210), so it does satisfy the orange juice constraint. The milk constraint is (6x + 3y\leq150), when (x = 0,y = 50), (6\times0+3\times50=150) which satisfies the milk constraint. So why did we get confused? Wait, the error was in the earlier step. The feasible region is defined by all (x\geq0,y\geq0), (9x + 3y\leq210) and (6x + 3y\leq150) Let's rewrite the inequalities: Inequality 1: (y\leq70 - 3x) Inequality 2: (y\leq50 - 2x) Since (50-2x\leq70 - 3x) (because ( - 2x+3x\leq70 - 50\Rightarrow x\leq20)), so for (x\leq20), the upper bound for (y) is (50 - 2x), and for (x>20), the upper bound for (y) is (70 - 3x) So the feasible region has vertices:

• ((0,0)): ( R = 0)
• ((0,50)): ( R=4\times0 + 3\times50=150) (check (9\times0+3\times50 = 150\leq210), (6\times0+3\times50=150\leq150), so this is a valid vertex)
• ((20,10)): ( R=4\times20+3\times10 = 80 + 30=110)
• ((\frac{70}{3},0)\approx(23.33,0)): ( R=4\times\frac{70}{3}+3\times0=\frac{280}{3}\approx93.33)

Now, we need to check if ((0,50)) satisfies both constraints:

• Orange juice: (9(0)+3(50)=150\leq210): yes
• Milk: (6(0)+3(50)=150\leq150): yes

But wait, when (x = 0,y = 50), the number of tropical smoothies is 0 and sport smoothies is 50. But let's check the problem statement again. The store makes two types of smoothies, but there is no constraint that they have to make both. However, let's check the orange juice and milk usage again.

Wait, the milk container is 150 - oz, and each sport smoothie uses 3 oz of milk, so 50 sport smoothies use (3\times50 = 150) oz of milk (which is exactly the container). Each sport smoothie uses 3 oz of orange juice, so 50 sport smoothies use (3\times50=150) oz of orange juice, and the orange juice container is 210 - oz, so that's within limit.

But wait, the revenue for 50 sport smoothies is (3\times50 = 150), and for 20 tropical and 10 sport smoothies is (4\times20+3\times10=80 + 30 = 110), for (\frac{70}{3}) tropical smoothies (about 23) and 0 sport smoothies is about (4\times23.33\approx93.33)

But this is a contradiction. Wait, no, the mistake is in the milk constraint. The milk used per tropical smoothie is 6 oz, per sport is 3 oz. So the milk constraint is (6x + 3y\leq150), which is correct. The orange juice constraint is (9x+3y\leq210), correct.

Wait, let's graph the inequalities:

For (3x + y\leq70) (orange juice), the line is (y=-3x + 70), intercepts at ((0,70)) and ((\frac{70}{3},0))

For (2x + y\leq50) (milk), the line is (y=-2x + 50), intercepts at ((0,50)) and ((25,0))

The feasible region is the area that is below both lines and in the first quadrant.

The intersection of (y=-3x + 70) and (y=-2x + 50) is at (x = 20,y = 10) as we found earlier.

The vertex ((0,50)) is on the milk line but above the orange juice line when (x = 0)? No, when (x = 0), (y=-3(0)+70 = 70) and (y=-2(0)+50 = 50). So the feasible region at (x = 0) is bounded by (y\leq50) (because (y=-2x + 50) is below (y=-3x + 70) when (x = 0), since (50<70)). So the correct feasible region's vertex at (x = 0) is ((0,50)), at (y = 0) is ((\frac{70}{3},0)), the intersection of the two lines is ((20,10)), and ((0,0))

Now, let's calculate the revenue at each vertex:

• ((0,0)): (R = 0)
• ((0,50)): (R=3\times50 = 150)
• ((20,10)): (R=4\times20+3\times10 = 110)
• ((\frac{70}{3},0)): (R=4\times\frac{70}{3}\approx93.33)

But this can't be right because the answer options include 110 and 150. Wait, maybe the problem is that the store makes two types of smoothies, and maybe there is an implicit constraint that they have to make at least some of each? But the problem doesn't say that.

Wait, no, let's re - read the problem: "a small organic food store makes two types of fruit smoothies: tropical and sport". It doesn't say they have to make both. But maybe the error is in my calculation. Wait, when (y = 50), (x = 0), let's check the milk: (6\times0+3\times50 = 150) (correct, milk container is 150 - oz). Orange juice: (9\times0+3\times50 = 150) (orange juice container is 210 - oz, correct). So why is the revenue 150? But the answer options include 150 and 110.

Wait, maybe the problem is that I misread the prices. The tropical smoothie is sold for $4, sport for $3. So 50 sport smoothies: (3\times50 = 150), 20 tropical and 10 sport: (4\times20+3\times10 = 110), 0 tropical and 0 sport: 0, 23 tropical and 0 sport: (4\times23 = 92) (since (\frac{70}{3}\approx23.33), so 23 tropical smoothies use (9\times23 = 207) oz of orange juice, (6\times23 = 138) oz of milk. Then remaining orange juice: (210 - 207 = 3) oz, remaining milk: (150 - 138 = 12) oz. But we can't make a fraction of a smoothie, so we have to use integer values?

Ah! The problem is that (x) and (y) must be non - negative integers (number of smoothies can't be a fraction). So my earlier mistake was not considering that (x) and (y) are integers.

So let's correct that.

We need to find integer values of (x) and (y) such that (9x + 3y\leq210) and (6x + 3y\leq150), and maximize (R = 4x+3y)

From (6x + 3y\leq150\Rightarrow2x + y\leq50\Rightarrow y\leq50 - 2x)

From (9x + 3y\leq210\Rightarrow3x + y\leq70\Rightarrow y\leq70 - 3x)

Since (y\leq50 - 2x) (because (50 - 2x\leq70 - 3x) for (x\leq20)), we can write (y = 50 - 2x + k), where (k\leq0) (since (y\leq50 - 2x)) and (y\geq0), (x\geq0)

We want to maximize (R = 4x+3y=4x + 3(50 - 2x + k)=4x+150-6x + 3k=150 - 2x+3k)

To maximize (R), we want to maximize ( - 2x+3k). Since (k\leq0), the maximum of ( - 2x+3k) occurs when (k = 0) and (x) is as small as possible (since the coefficient of (x) is negative). But when (k = 0), (y = 50 - 2x)

We also have (9x + 3y\leq210). Substitute (y = 50 - 2x) into the orange juice constraint:

(9x+3(50 - 2x)\leq210)

(9x + 150-6x\leq210)

(3x\leq60)

(x\leq20)

Since (x) is a non - negative integer, (x) can be from 0 to 20.

Now, (R = 4x+3(50 - 2x)=4x + 150-6x=150 - 2x)

To maximize (R), we need to minimize (x). The minimum value of (x) is 0.

When (x = 0), (y = 50 - 2\times0 = 50)

Check orange juice: (9\times0+3\times50 = 150\leq210) (correct)

Milk: (6\times0+3\times50 = 150\leq150) (correct)

Revenue: (4\times0+3\times50 = 150)

But wait, if (x = 0,y = 50), that's 50 sport smoothies. But let's check if we can make more revenue by having some tropical smoothies.

Let's try (x = 1), then (y\leq50 - 2\times1 = 48)

Check orange juice: (9\times1+3\times48=9 + 144 = 153\leq210)

Revenue: (4\times1+3\times48=4 + 144 = 148) (less than 150)

(x = 2), (y\leq46)