question 4 an electronics company assembles two types of tvs: plasma and lcd. a plasma tv costs $400 to…

question 4 an electronics company assembles two types of tvs: plasma and lcd. a plasma tv costs $400 to assemble and takes 40 hours of labor. the lcd tv costs $250 and requires 30 hours of labor. the company has $20 000 in capital, and 2160 hours of labor available for assembly. what is the maximum number of tvs the electronics company can assemble?

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of plasma - TVs and $y$ be the number of LCD - TVs.

Step2: Set up constraints

Cost constraint: $400x + 250y\leq20000$ (capital constraint), which simplifies to $8x + 5y\leq400$. Labor - hour constraint: $30x+40y\leq2160$, which simplifies to $3x + 4y\leq216$. Also, $x\geq0,y\geq0$ (non - negativity constraints).

Step3: Express the objective function

We want to maximize $N=x + y$ (total number of TVs).

Step4: Solve the system of inequalities

First, rewrite the inequalities as equations to find the corner points: For $8x + 5y=400$, when $x = 0$, $y = 80$; when $y = 0$, $x = 50$. For $3x + 4y=216$, when $x = 0$, $y = 54$; when $y = 0$, $x = 72$. Now, solve the system of equations $\begin{cases}8x + 5y=400\3x + 4y=216\end{cases}$ Multiply the first equation by 4 and the second by 5: $32x+20y = 1600$ and $15x+20y=1080$ Subtract the second equation from the first: $(32x + 20y)-(15x + 20y)=1600 - 1080$ $32x-15x=520$ $17x=520$ $x=\frac{520}{17}\approx30.59$ Substitute $x$ into $8x + 5y=400$: $8\times\frac{520}{17}+5y=400$ $\frac{4160}{17}+5y=400$ $5y=400-\frac{4160}{17}$ $5y=\frac{6800 - 4160}{17}=\frac{2640}{17}$ $y=\frac{2640}{85}\approx31.06$ The corner points of the feasible region are $(0,0),(0, 54),(50,0)$ and the intersection point of the two non - trivial lines. Evaluate the objective function $N=x + y$ at each corner point:

At $(0,0)$: $N=0+0 = 0$
At $(0,54)$: $N=0 + 54=54$
At $(50,0)$: $N=50+0 = 50$
For the intersection point $x=\frac{520}{17}\approx30.59,y=\frac{2640}{85}\approx31.06$, $N=\frac{520}{17}+\frac{2640}{85}=\frac{2600 + 2640}{85}=\frac{5240}{85}\approx61.65$

Since we are dealing with whole number of TVs, we consider the integer values in the feasible region. We can also test integer points close to the non - integer intersection point. By checking integer points in the feasible region, we find that when $x = 30,y = 31$ $8x+5y=8\times30 + 5\times31=240+155 = 395\leq400$ $3x + 4y=3\times30+4\times31=90 + 124=214\leq216$ $N=x + y=30 + 31=61$