Respuesta :
The derivative of a function indicates how sensitive the function is to changes as the input variable value changes
The revenue, profit, maximum profit and price are;
(a) Her revenue on the day is 300 - 10·x - 4·x²
(b) Her profit function, P(x) = 120 + 14·x - 4·x²
(c) The price she should set the necklace is $11.75
(d) The maximum profit made is $132.25
The reasons the above values are correct are as follows:
The given parameters are:
Cost of material for each necklace =$6
The number of necklace she sells a day = 30
Selling price for each necklace when she sells 30 necklaces = $10
Number of sales lost for every dollar increase = 4 sales a day
(a) When the price is increased by x dollars, then we have;
New price = 10 + x
New cost of daily production, C = 6·(30 - 4·x) dollars
The number of sales for the day = 30 - 4·x
Her revenue for the day, R = (10 + x)·(30 - 4·x) = 300 - 10·x - 4·x²
(b) Her profit, P = Her daily revenue, R - Her daily production cost, C
Her cost function, P(x) = R - C
∴ P(x) = 300 - 10·x - 4·x² - 6·(30 - 4·x) = 120 + 14·x - 4·x²
Her profit as a function of the number of x dollar increase in price, is therefore, P(x) = 120 + 14·x - 4·x²
(c) To maximize profit, the derivative of the profit function is found and equated to zero as follows;
[tex]\dfrac{dP(x)}{dx} = \dfrac{d}{dx} \left(120 + 14\cdot x - 4 \cdot x^2\right) = 14 - 8\cdot x[/tex]
x = 14/8 = 1.75
Therefore, the price she should set the necklace = 10 + 1.75 = 11.75
The price she should set the necklace = $11.75
(d) The maximum profit [tex]P(x)_{max}[/tex] = P(1.75) = 120 + 14×1.75 - 4×1.75² = 132.25
Her maximum profit, [tex]P(x)_{max}[/tex] = $132.25
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