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Svetach [21]
3 years ago
5

1. 5 Burgers and 3 orders of fries cost $34. 4 Burgers and 4 orders of fries cost $32. How much does each burger and each order

of fries cost?
2. A pile of 44 coins is made up of pennies, nickels, and dimes. If the pile is worth $3.07 and there are twice as many nickels as pennies, how many of each coin does the pile contain?
Mathematics
1 answer:
Elena L [17]3 years ago
7 0

Answer:

1.

a burger costs $5

an order of fries costs $3

2.

7 pennies

14 nickels

23 dimes

Step-by-step explanation:

x = price of a burger

y = price of a order of fries

5x + 3y = 34

4x + 4y = 32

=>

x + y = 8

x = 8-y

5×(8-y) + 3y = 34

40 - 5y + 3y = 34

-2y = -6

y = $3

x = 8 - 3 = $5

x = number of pennies

y = number of nickels

z = number of dimes

0.01×x + 0.05×y + 0.1×z = 3.07

y = 2x

x + y + z = 44

x + 2x + z = 44

3x + z = 44

z = 44 - 3x

0.01×x + 0.05×2x + 0.1×z = 3.07

0.11×x + 0.1×z = 3.07

0.11×x + 0.1×(44-3x) = 3.07

11x + 10×(44-3x) = 307

11x + 440 - 30x = 307

-19x = -133

x = 7

y = 2×7 = 14

z = 44 - 3×7 = 44 - 21 = 23

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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.

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3(2x - 1) = 1/2 (4x - 2) + 2<br> What would it be?
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The quotient of (x4 + 5x3 – 3x – 15) and a polynomial is (x3 – 3). What is the polynomial?
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