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andre [41]
3 years ago
11

A padaria Beira Baixa vem perdendo dinheiro com a sua política de produção de venda do pãozinho francês. Atualmente, a cada dia

é feita uma quantidade igual à procura do dia anterior. O proprietário, Sr. Joaquim, recolheu os seguintes dados:
Lucro na venda= $ 0,20 por unidade

Perda no pão não vendido = $ 0,05

(os pães não vendidos são moídos e vendidos como farinha de rosca: cada pão equivale a $0,15 de farinha de rosca)

Distribuição de frequência da demanda diária. Quantos pães devem ser fabricados diariamente para maximizar o lucro médio do Sr. Joaquim


1000 0,10

1100 0,10

1200 0,10

1300 0,15

1400 0,20

1500 0,15

1600 0,10

1700 0,05

1800 0,05
Mathematics
1 answer:
Finger [1]3 years ago
8 0

Answer:

The number of loaves to be made daily to maximize Mr. Joaquim's average profit is 1400.

O número de pães a serem feitos diariamente para maximizar o lucro médio do Sr. Joaquim é 1400.

Step-by-step explanation:

The table gives the daily demand of bread and the average profit that corresponds to each demand.

The daily demand values recorded ranged from 1000 to 1800.

And the corresponding average profit for the demands for each daily demand provided showed a steady increase from $0.10 as at a daily demand of 1000, through $0.15, then peaking at $0.20 at a daily demand of 1400.

After this point, the average profit begins to decline again as we move further away from the peak obtained at a daily demand of 1400.

Hence, it is evident that the daily demand that corresponds to the maximum average profit is 1400.

A daily demand of 1400 is where the peak average profit of $0.20 is obtained.

Any other demand lower or higher than 1400 corresponds to a departure from the peak profit obtained at that point according to the table of daily demand and average profits provided in the question.

In portugese/Em português

A tabela fornece a demanda diária de pão e o lucro médio que corresponde a cada demanda.

Os valores diários de demanda registrados variaram de 1000 a 1800.

E o lucro médio correspondente às demandas de cada demanda diária fornecida mostrou um aumento constante de US $ 0,10, com uma demanda diária de 1000, passando por US $ 0,15, atingindo um pico de US $ 0,20 com uma demanda diária de 1400.

Após esse ponto, o lucro médio começa a declinar novamente à medida que nos afastamos do pico obtido com uma demanda diária de 1400.

Portanto, é evidente que a demanda diária que corresponde ao lucro médio máximo é de 1.400.

Uma demanda diária de 1400 é onde é obtido o lucro médio máximo de US $ 0,20.

Qualquer outra demanda menor ou maior que 1400 corresponde a um afastamento do pico de lucro obtido nesse momento, de acordo com a tabela de demanda diária e lucros médios fornecidos na pergunta.

Hope this Helps!!!

Espero que isto ajude!!!

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A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)=74,000+80x
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c) 3225 set, $272687.5 profit, $192.5 per set

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The maximum revenue is at R'(x) =0

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But we need to compute R'(x) = 0:

300 - x/15 = 0

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R(4500) = 300 (4500) - (4500)²/30 = $675000  

B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000

P(x) = -x²/30 + 220x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 220= -x/15 + 220

But we need to compute P'(x) = 0:

-x/15 + 220 = 0

x/15 = 220

x = 3300

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:

P(3300) =  -(3300)²/30 + 220(3300) - 74000 = $289000  

The price for each set is:

p(3300) = 300 -3300/30 = $190 per set

c) The new cost is:

C(x) = 74000 + 80x + 5x = 74000 + 85x

Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000

P(x) = -x²/30 + 215x - 74000

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But we need to compute P'(x) = 0:

-x/15 + 215 = 0

x/15 = 215

x = 3225

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:

P(3225) =  -(3225)²/30 + 215(3225) - 74000 = $272687.5

The money to be charge for each set is:

p(x) = 300 - 3225/30 = $192.5 per set

When taxed $5, the maximum profit is $272687.5

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