A custom guitar shop sells two types of guitars: electric and acoustic. The creation process of each guitar is broken into three parts: building the base guitar, applying the finis... A custom guitar shop sells two types of guitars: electric and acoustic. The creation process of each guitar is broken into three parts: building the base guitar, applying the finish/paint, and adding accessories. The time requirements (in days) and the total days available for each phase of the creation process are shown in the below table. If they profit $400 from each electric guitar and $300 from each acoustic guitar, how many of each guitar type should be made to maximize the shop's profit? What is the shop's maximum profit (in dollars)? The maximum profit is $_____ when _____ electric guitars and _____ acoustic guitars are made. Determine if the shop has any leftover time (in days) for each task, entering 0 if no time is leftover. There are _____ days leftover for building base guitars, _____ days leftover for applying finish/paint, and _____ days leftover for adding accessories. Read more

Steps to Solve

1. Define variables

Let $x$ be the number of electric guitars and $y$ be the number of acoustic guitars.

2. Objective function

The profit function to maximize is $P = 400x + 300y$.

3. Constraints

Based on the table, we have the following constraints:

$6x + 4y \le 516$ (Base guitar creation)

$2x + 2y \le 218$ (Applying the finish/paint)

$2x + 5y \le 470$ (Adding accessories)

Also, $x \ge 0$ and $y \ge 0$ since we cannot produce a negative number of guitars.

4. Simplify the constraints

We can simplify the constraints:

$6x + 4y \le 516 \implies 3x + 2y \le 258$

$2x + 2y \le 218 \implies x + y \le 109$

$2x + 5y \le 470$

5. Find the corner points of the feasible region

First, let's find the intersection of $3x + 2y = 258$ and $x + y = 109$. Multiply the second equation by 2 to get $2x + 2y = 218$. Subtracting this from the first equation gives:

$(3x + 2y) - (2x + 2y) = 258 - 218 \implies x = 40$

Then $40 + y = 109 \implies y = 69$.

So, one corner point is $(40, 69)$.

Next, let's find the intersection of $x + y = 109$ and $2x + 5y = 470$. Multiply the first equation by 2 to get $2x + 2y = 218$. Subtracting this from the second equation gives:

$(2x + 5y) - (2x + 2y) = 470 - 218 \implies 3y = 252 \implies y = 84$

Then $x + 84 = 109 \implies x = 25$.

So, another corner point is $(25, 84)$.

Next, we need to find the intersection of $3x+2y=258$ and $2x+5y=470$.

Multiply the first equation by 5 and the second by 2:

$15x + 10y = 1290$ and $4x + 10y = 940$.

Subtract the second from the first:

$11x = 350 \implies x = \frac{350}{11} \approx 31.82$

$2(\frac{350}{11}) + 5y = 470 \implies 5y = 470 - \frac{700}{11} \implies 5y = \frac{5170 - 700}{11} = \frac{4470}{11} \implies y = \frac{894}{11} \approx 81.27$

Since we are dealing with whole guitars we need to consider integer values close to this intersection when maximizing or minimizing the objective function. However in this problem we would get a lower profit by rounding to adjacent integers.

Finally we also have the points $(0,0)$, $(0, 94)$ where $x=0$ and $2(0) + 5y = 470 \implies y = 94$, and $(86, 0)$ where $y = 0$ and $3x + 2(0) = 258 \implies x = 86$, and $(109,0)$ where $x = 109$ and $y=0$, but $3(109) > 258$ so this solution is invalid.

6. Evaluate the objective function at the corner points

$P(0,0) = 400(0) + 300(0) = 0$

$P(86,0) = 400(86) + 300(0) = 34400$

$P(0,94) = 400(0) + 300(94) = 28200$

$P(40,69) = 400(40) + 300(69) = 16000 + 20700 = 36700$

$P(25,84) = 400(25) + 300(84) = 10000 + 25200 = 35200$

$P(31.82, 81.27)$ is not accurate since they must be integers, and $40, 69$ is better.

7. Determine the maximum profit

The maximum profit is $36,700 when 40 electric guitars and 69 acoustic guitars are made.

8. Calculate leftover time

For base guitar creation: $6(40) + 4(69) = 240 + 276 = 516$. Leftover time: $516 - 516 = 0$ days.

For applying finish/paint: $2(40) + 2(69) = 80 + 138 = 218$. Leftover time: $218 - 218 = 0$ days.

For adding accessories: $2(40) + 5(69) = 80 + 345 = 425$. Leftover time: $470 - 425 = 45$ days.