Question: A company produces two types of calculators: basic and scientific. A basic calculator requires 2 hours of assembly and 1 hour of testing. A scientific calculator requires 3 hours of assembly and 2 hours of testing.
The company has 120 assembly hours and 70 testing hours available each week. The profit is 25 for each basic calculator and 40 for each scientific calculator.
How many of each type should the company produce to maximize its weekly profit?
1. **State the problem:**
We want to find how many basic calculators ($x$) and scientific calculators ($y$) the company should produce to maximize profit given constraints on assembly and testing hours.
2. **Define variables:**
Let $x$ = number of basic calculators produced.
Let $y$ = number of scientific calculators produced.
3. **Write the constraints:**
Assembly hours: $2x + 3y \leq 120$
Testing hours: $1x + 2y \leq 70$
Also, $x \geq 0$, $y \geq 0$ since production quantities cannot be negative.
4. **Write the profit function to maximize:**
$$P = 25x + 40y$$
5. **Find the feasible region by solving constraints:**
From assembly: $2x + 3y \leq 120$
From testing: $x + 2y \leq 70$
6. **Find intersection points of constraints:**
Solve system:
$$\begin{cases} 2x + 3y = 120 \\ x + 2y = 70 \end{cases}$$
Multiply second equation by 2:
$$2x + 4y = 140$$
Subtract first equation:
$$2x + 4y - (2x + 3y) = 140 - 120$$
$$y = 20$$
Substitute $y=20$ into $x + 2y = 70$:
$$x + 2(20) = 70$$
$$x + 40 = 70$$
$$x = 30$$
7. **Evaluate profit at corner points:**
- At $(0,0)$: $P=25(0)+40(0)=0$
- At $(0,35)$ from $x+2y=70$ when $x=0$: $P=25(0)+40(35)=1400$
- At $(60,0)$ from $2x+3y=120$ when $y=0$: $P=25(60)+40(0)=1500$
- At $(30,20)$ intersection: $P=25(30)+40(20)=750+800=1550$
8. **Conclusion:**
Maximum profit is $1550$ when producing $30$ basic and $20$ scientific calculators.
**Final answer:**
Produce $\boxed{30}$ basic and $\boxed{20}$ scientific calculators to maximize weekly profit of $1550$.