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vredina [299]
3 years ago
14

The volleyball team raised $350 from its fundraiser. It wants to buy jerseys and volleyballs. The jerseys cost $12 each, and the

team needs 22 jerseys. Volleyballs cost $15 each. What is the maximum number of volleyballs the team can buy and still have money left? Write your answer and explain the steps you took. WILL GIVE BRAIKNLIEST IF RIGHT!!!!!!!!
Mathematics
2 answers:
Anna71 [15]3 years ago
7 0

Answer:

6

Step-by-step explanation:

Sphinxa [80]3 years ago
7 0

Answer:

The team can buy 5 volleyballs at maximum with no money left over.

Step-by-step explanation:

First, you need to find the money that the jerseys used:

12 × 22 = $264

Next, you have to subtract that from the total amount:

350 - 264 = $86

Finally, you have to see how many volleyballs you can buy:

If volleyballs cost $15 each then, 86 ÷ 15 = 5.73333333(Infinite)

But you can not buy part of a volleyball so you would have to buy 5 volleyballs. Then you would only have .7333333(Infinite) money left over, but that wouldn't be good for anything.

Is this a good answer????

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A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)equals72 com
solmaris [256]

Answer:

Part (A)

  • 1. Maximum revenue: $450,000

Part (B)

  • 2. Maximum protit: $192,500
  • 3. Production level: 2,300 television sets
  • 4. Price: $185 per television set

Part (C)

  • 5. Number of sets: 2,260 television sets.
  • 6. Maximum profit: $183,800
  • 7. Price: $187 per television set.

Explanation:

<u>0. Write the monthly cost and​ price-demand equations correctly:</u>

Cost:

      C(x)=72,000+70x

Price-demand:

     

      p(x)=300-\dfrac{x}{20}

Domain:

        0\leq x\leq 6000

<em>1. Part (A) Find the maximum revenue</em>

Revenue = price × quantity

Revenue = R(x)

           R(x)=\bigg(300-\dfrac{x}{20}\bigg)\cdot x

Simplify

      R(x)=300x-\dfrac{x^2}{20}

A local maximum (or minimum) is reached when the first derivative, R'(x), equals 0.

         R'(x)=300-\dfrac{x}{10}

Solve for R'(x)=0

      300-\dfrac{x}{10}=0

       3000-x=0\\\\x=3000

Is this a maximum or a minimum? Since the coefficient of the quadratic term of R(x) is negative, it is a parabola that opens downward, meaning that its vertex is a maximum.

Hence, the maximum revenue is obtained when the production level is 3,000 units.

And it is calculated by subsituting x = 3,000 in the equation for R(x):

  • R(3,000) = 300(3,000) - (3000)² / 20 = $450,000

Hence, the maximum revenue is $450,000

<em>2. Part ​(B) Find the maximum​ profit, the production level that will realize the maximum​ profit, and the price the company should charge for each television set. </em>

i) Profit(x) = Revenue(x) - Cost(x)

  • Profit (x) = R(x) - C(x)

       Profit(x)=300x-\dfrac{x^2}{20}-\big(72,000+70x\big)

       Profit(x)=230x-\dfrac{x^2}{20}-72,000\\\\\\Profit(x)=-\dfrac{x^2}{20}+230x-72,000

ii) Find the first derivative and equal to 0 (it will be a maximum because the quadratic function is a parabola that opens downward)

  • Profit' (x) = -x/10 + 230
  • -x/10 + 230 = 0
  • -x + 2,300 = 0
  • x = 2,300

Thus, the production level that will realize the maximum profit is 2,300 units.

iii) Find the maximum profit.

You must substitute x = 2,300 into the equation for the profit:

  • Profit(2,300) = - (2,300)²/20 + 230(2,300) - 72,000 = 192,500

Hence, the maximum profit is $192,500

iv) Find the price the company should charge for each television set:

Use the price-demand equation:

  • p(x) = 300 - x/20
  • p(2,300) = 300 - 2,300 / 20
  • p(2,300) = 185

Therefore, the company should charge a price os $185 for every television set.

<em>3. ​Part (C) If the government decides to tax the company ​$4 for each set it​ produces, how many sets should the company manufacture each month to maximize its​ profit? What is the maximum​ profit? What should the company charge for each​ set?</em>

i) Now you must subtract the $4  tax for each television set, this is 4x from the profit equation.

The new profit equation will be:

  • Profit(x) = -x² / 20 + 230x - 4x - 72,000

  • Profit(x) = -x² / 20 + 226x - 72,000

ii) Find the first derivative and make it equal to 0:

  • Profit'(x) = -x/10 + 226 = 0
  • -x/10 + 226 = 0
  • -x + 2,260 = 0
  • x = 2,260

Then, the new maximum profit is reached when the production level is 2,260 units.

iii) Find the maximum profit by substituting x = 2,260 into the profit equation:

  • Profit (2,260) = -(2,260)² / 20 + 226(2,260) - 72,000
  • Profit (2,260) = 183,800

Hence, the maximum profit, if the government decides to tax the company $4 for each set it produces would be $183,800

iv) Find the price the company should charge for each set.

Substitute the number of units, 2,260, into the equation for the price:

  • p(2,260) = 300 - 2,260/20
  • p(2,260) = 187.

That is, the company should charge $187 per television set.

7 0
2 years ago
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