Question

During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for $10 each and his sales averaged 20 per day. When he increased the price by $1, he found that the average decreased by two sales per day. If the material for each necklace costs Terry $6, what should the selling price be to maximize his profit?

Answer 1

Answer:

The selling price in other to maximize his profit is $13

Step-by-step explanation:

In the above question we are given the following information:

Cost of material per necklace = $6

Firstly, terry sold 20 necklaces per day

= $10 each

Later he increased he increased the prices by 1 dollar and the number of necklaces he sold reduced by 2

Mathematically

18 necklaces = $11 each

Step 1

We find the Cost function C(x)

Let's assume that x = number of necklaces sold

If each material cost $6 , then

C(x) = 6 × x = 6x

Step 2

P(Profit) = R(x) - C(x)

R(x) = Revenue

Where Revenue = x × p(x)

Since p(20) = 10 and p(18) = 11

p(x) = -1/2x + 20

P(Profit) = x ( -1/2x + 20) - C(x)

C(x) = 6x

P = x(-1/2x + 20) - 6x

P = -1/2x² + 20x - 6x

P = -1/2x² + 14x

Step 3

We maximise the profit by differentiating P

P = Profit

P = -1/2x² + 14x

We differentiate P to find x

∆P/∆x = dp/dx = -x + 14

-x + 14 = 0

-x = -14

x = 14

Hence, we substitute 14 for x in the price function

p(x) = - 1/2x + 20

since , x = 14

p(14) = - 1/2 × 14 + 20

= -7 + 20

= $13

Therefore, the selling price function to maximize his profit is $13

Answer 2

Above question the given data:  

• Cost of material per necklace = $6
• Terry sold 20 necklaces per day = $10 each
• Price increase by 1 dollar
• Number of necklaces  sold reduced by 2  

1.Cost function C(x)

Let's assume that x = number of necklaces sold

If each material cost $6 , then

C(x) = 6 × x

C(x) = 6x  

2.P(Profit) = R(x) - C(x)

R(x) = Revenue ,Where Revenue = x × p(x)

Given data:

p(20) = 10

p(18) = 11

p(x) = -1/2x + 20

P(Profit) = R(x) - C(x)

P(Profit) = x ( -1/2x + 20) - C(x)

P = x(-1/2x + 20) - 6x

P = -1/2x² + 20x - 6x

P = -1/2x² + 14x

 

3.Maximise Profit

P = Profit  

P = -1/2x² + 14x

We differentiate P to find x

∆P/∆x = dp/dx = -x + 14

-x + 14 = 0

-x = -14

x = 14

Now, we will substitute 14 for x in the price function

Now ,p(x) = - 1/2x + 20

since , x = 14

p(14) = - 1/2 × 14 + 20  

p(14)= -7 + 20  

p(14)= $13

Thus, the selling price function to maximize his profit is $13.

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