a tire company is selling two different tread patterns of tires. tire x sells for $75.00 and tire y sells…

a tire company is selling two different tread patterns of tires. tire x sells for $75.00 and tire y sells for $85.00. three times the number of tire y sold must be less than or equal to twice the number of x tires sold. the company has at most 300 tires to sell. what is the maximum revenue that the company can make?\n$13,500\n$22,500\n$23,700\n$25,500

a tire company is selling two different tread patterns of tires. tire x sells for $75.00 and tire y sells for $85.00. three times the number of tire y sold must be less than or equal to twice the number of x tires sold. the company has at most 300 tires to sell. what is the maximum revenue that the company can make?\n$13,500\n$22,500\n$23,700\n$25,500

Answer

Explanation:

Step1: Define the revenue function

The revenue $R$ from selling $x$ tires of type - $x$ and $y$ tires of type - $y$ is given by $R = 75x+85y$.

Step2: Identify the constraints

We have two constraints:

  1. $3y\leq2x$ or $3y - 2x\leq0$.
  2. $x + y\leq300$ and $x\geq0,y\geq0$ (non - negative number of tires). The intersection points of the boundary lines of the constraints are $(0,0),(300,0),(0,300),(180,120)$ (from solving the system of equations $\begin{cases}x + y=300\3y = 2x\end{cases}$).

Step3: Evaluate the revenue function at the intersection points

  • At $(0,0)$: $R=75\times0 + 85\times0=0$.
  • At $(300,0)$: $R=75\times300+85\times0 = 22500$.
  • At $(0,300)$: $R=75\times0+85\times300 = 25500$.
  • At $(180,120)$: $R=75\times180+85\times120=75\times180 + 85\times120=13500+10200=23700$.

Answer:

$25500$ (D. $$25,500$)