Question

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5. It costs $800 to manufacture a certain model of personal computer Overhead and other fixed costs to the company
are $2000 per week. The wholesale price of a computer is $1500 but, as an incentive, the company will reduce the price
of every computer by an additional $10 for each computer purchased in excess of 10 (Thus if 13 computers are
purchased, each will cost $1470.) Express the company weekly profit as a function of the number of computers sold.​

Answer 1

Question :-

It costs $800 to manufacture a certain model of personal computer Overhead and other fixed costs to the company

are $2000 per week. The wholesale price of a computer is $1500 but, as an incentive, the company will reduce the price of every computer by an additional $10 for each computer purchased in excess of 10 (Thus if 13 computers arepurchased, each will cost $1470.) Express the company weekly profit as a function of the number of computers sold.

Answer

$\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}}\end{gathered}$

Let number of additional computer be 'x'.

Cost Price of '10 + x' computers = 2000 + 800(10 + x)

Selling Price per Computer = 1500 - 10x

$\begin{gathered}\bf\red{According\:to\:the\:question,} \\ \end{gathered}$

$\bf \: ⟼ Number \: of \: computers \: = x + 10$

$\bf \: ⟼ Price \: per \: computer = 1500 - 10x$

Revenue proceed, R

$\bf \: ⟼ R = (10 + x)(1500 - 10x)$

$\bf \: ⟼ R = 15000 - 100x + 1500x - 10 {x}^{2}$

$\bf \: ⟼ R = 15000 + 1400x - 10 {x}^{2} \: ⟼ \: (1)$

$\bf \: ⟼ Profit = 15000 + 1400x - {10x}^{2} - 2000 - 800(10 + x)$

$\bf \: ⟼ P = 7000 + 600x - {10x}^{2}$

Differentiate w. r. t. x, we get

$\bf \: ⟼ \dfrac{dP}{dx} = 0 + 600 - 20x \: ⟼ \: (2)$

For maxima or minima,

$\bf \: ⟼ \dfrac{dP}{dx} = 0$

$\bf \: ⟼ 600 - 20x = 0$

$\bf \: ⟼ x = 30$

Differentiate equation (2), w. r. t. x, we get

$\bf \: ⟼ \dfrac{ {d}^{2} P}{d {x}^{2} } = - 20 < 0$

$\bf \: ⟼ Profit \: is \: maximum \: when \: 40 \: units \: are \: purchases.$

$\begin{gathered}\bf\red{So,}\end{gathered}$

$\bf \: ⟼Profit( P) = 7000 + 600x - {10x}^{2}$