a store sells two brands of tvs. brand a sells at least twice as much as brand b. the store must carry at…

a store sells two brands of tvs. brand a sells at least twice as much as brand b. the store must carry at least 10 of brand b, and has room for no more than 90 tvs. the store makes a profit of $100 on brand a tvs and a profit of $75 on brand b tvs. what combinations of tvs would maximize the stores profit?

a store sells two brands of tvs. brand a sells at least twice as much as brand b. the store must carry at least 10 of brand b, and has room for no more than 90 tvs. the store makes a profit of $100 on brand a tvs and a profit of $75 on brand b tvs. what combinations of tvs would maximize the stores profit?

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of Brand A TVs and $y$ be the number of Brand B TVs.

Step2: Write constraints

  1. $x\geq2y$ (Brand A sells at least twice as much as brand B)
  2. $y\geq10$ (At least 10 of brand B)
  3. $x + y\leq90$ (No more than 90 TVs in total)
  4. $x\geq0,y\geq0$ (Non - negative number of TVs)

Step3: Write the profit function

The profit function $P = 100x+75y$

Step4: Find the corner points of the feasible region

  1. Intersection of $y = 10$ and $x=2y$: Substitute $y = 10$ into $x = 2y$, we get $x=20$. So the point is $(20,10)$.
  2. Intersection of $y = 10$ and $x + y=90$: Substitute $y = 10$ into $x + y=90$, we get $x=80$. So the point is $(80,10)$.
  3. Solve the system of equations $\begin{cases}x=2y\x + y=90\end{cases}$. Substitute $x = 2y$ into $x + y=90$, we have $2y+y=90$, $3y=90$, $y = 30$, and $x=60$. So the point is $(60,30)$.

Step5: Evaluate the profit function at corner points

  1. At $(20,10)$: $P=100\times20 + 75\times10=2000 + 750=2750$
  2. At $(80,10)$: $P=100\times80+75\times10=8000 + 750=8750$
  3. At $(60,30)$: $P=100\times60+75\times30=6000+2250=8250$

Answer:

The store should sell 80 Brand A TVs and 10 Brand B TVs to maximize profit.