Question

Mrs.mcgee makes craft jewelry to sell at a seasonal craft makes pins and pin takes 1 hour to make and sell profit of pairs of earrings take 2 hours to make, but she gets a profit of $ likes to have a variety, so she wants to have at least as many pins as pairs of also knows that she has approximately 40 hours for creating jewelry between now and the start of the also knows that the craft show vendor wants sellers to have more than 20 items on display at the beginning of the show.assuming she sells all her inventory,how many of pins and earrings pairs should the woman make to maximilian her profit?

Answer 1

Mrs. McGee should make 15 pins and 10 pairs of earrings to maximize her profit.

Step-by-step explanation:

To maximize profit, Mrs. McGee needs to balance the time it takes to create each item and the profit generated from their sale. Let \(P\) represent the number of pins and \(E\) represent the number of pairs of earrings. The time constraint is given as
`\(1P + 2E \leq 40`) hours, and the requirement for more than 20 items can be expressed as (P + E > 20).

Considering the profit aspect, each pin yields a profit of (1) and each pair of earrings yields a profit of (2). Therefore, the objective function for profit is (Z = P + 2E).

Now, with the constraints and objective function, we can set up the following linear programming problem:

`\text{Maximize:} \quad & Z = P + 2E \\`

`\text{Subject to:} \quad & P + 2E \leq 40 \\`

`& P + E > 20 \\& P, E \geq 0\end{align*}`

By solving this system of inequalities, we find the optimal values for (P) and (E), which are (15) pins and (10) pairs of earrings. This combination ensures Mrs. McGee meets the time constraint, the item display requirement, and maximizes her profit.

Answer 2

To maximize her profit at the craft show, Mrs. McGee should make 20 pins and 10 pairs of earrings.

Step-by-step explanation:

Define variables:

p: Number of pins

e: Number of pairs of earrings

t: Total time available (40 hours)

m: Minimum number of items (20)

pp: Profit per pin ($8)

pe: Profit per pair of earrings ($20)

Constraints:

Time constraint: p + 2e <= t => p + 2e <= 40

Minimum items constraint: p + e >= m => p + e >= 20

Variety constraint: p >= e

Objective function:

Maximize total profit: P(p, e) = pp * p + pe * e

Method:

Convert the constraints into inequalities:

Time: 2e <= 40 - p => e <= 20 - p/2

Minimum items: e >= 20 - p

Variety: p >= e

Substitute the time constraint into the objective function:

P(p) = pp * p + pe * (20 - p/2)

Differentiate P(p) and set it equal to zero to find the optimal p:

dP(p)/dp = pp - pe/2 = 0

Solving for p: p = 2e

Substitute p = 2e into the constraints:

e <= 20 - 2e/2 => e <= 10

e >= 20 - 2e => e >= 6.67

2e >= e => e <= 2e

The feasible range for e is 6.67 <= e <= 10.

We want to maximize P(p), so we choose the highest value of e within the feasible range: e = 10.

Calculate the corresponding number of pins: p = 2e = 20.

Results:

Number of pins (p): 20

Number of pairs of earrings (e): 10

Total profit: P(20, 10) = 8 * 20 + 20 * 10 = $360

Therefore, Mrs. McGee should make 20 pins and 10 pairs of earrings to maximize her profit at the craft show.