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

a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie uses 9 oz of orange juice and 6 oz of milk. each sport smoothie uses 3 oz of orange juice and 3 oz of milk. the store buys the orange juice in a 210 - oz container and milk in a 150 - oz container. the store sells the tropical smoothies for $4 and sells the sport smoothie for $3. what is the maximum in smoothie sales the store can make from 1 container each of orange juice and milk?\n$100\n$110\n$150\n$280

a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie uses 9 oz of orange juice and 6 oz of milk. each sport smoothie uses 3 oz of orange juice and 3 oz of milk. the store buys the orange juice in a 210 - oz container and milk in a 150 - oz container. the store sells the tropical smoothies for $4 and sells the sport smoothie for $3. what is the maximum in smoothie sales the store can make from 1 container each of orange juice and milk?\n$100\n$110\n$150\n$280

Answer

Explanation:

Step1: Set up variables

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

Step2: Write constraints based on orange - juice

The amount of orange - juice used gives the constraint: $9x + 3y\leqslant210$, which simplifies to $3x + y\leqslant70$.

Step3: Write constraints based on milk

The amount of milk used gives the constraint: $6x+3y\leqslant150$, which simplifies to $2x + y\leqslant50$. Also, $x\geqslant0,y\geqslant0$ since the number of smoothies cannot be negative.

Step4: Write the objective function

The profit function $P$ that we want to maximize is $P = 4x+3y$.

Step5: Solve the system of inequalities for corner - points

First, solve the system $\begin{cases}3x + y=70\2x + y=50\end{cases}$. Subtract the second equation from the first: $(3x + y)-(2x + y)=70 - 50$, which gives $x = 20$. Substitute $x = 20$ into $2x + y=50$, we get $2\times20+y=50$, so $y = 10$. The corner - points of the feasible region are found by considering the intersection of the boundary lines and the non - negativity constraints: Intersection of $x = 0$ and $3x + y=70$ gives $(0,70)$; $P(0,70)=4\times0 + 3\times70=210$. Intersection of $x = 0$ and $2x + y=50$ gives $(0,50)$; $P(0,50)=4\times0+3\times50 = 150$. Intersection of $y = 0$ and $3x + y=70$ gives $( \frac{70}{3},0)$; $P(\frac{70}{3},0)=4\times\frac{70}{3}+3\times0=\frac{280}{3}\approx93.33$. Intersection of $y = 0$ and $2x + y=50$ gives $(25,0)$; $P(25,0)=4\times25+3\times0 = 100$. Intersection of $3x + y=70$ and $2x + y=50$ gives $(20,10)$; $P(20,10)=4\times20+3\times10=80 + 30=110$.

Answer:

$110$