At what price should the company sell the two systems in order to maximise profit?
A telephone company is planning to introduce two new types of executive communication systems. It is estimated that if the first type is priced at x hundred dollars per system and the second at y hundred dollars per system, 42-6x+7y consumers will buy the first type and 66+4x-7y will buy the second type. If the cost of manufacturing both types is $2000 per system, how should the company price the systems to generate the largest possible profit? Round your answers to the nearest cent.
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The price of the first system is `x` x 100$ and the price of the second is `y` x 100$. Each costs $2000 ( = 20 x 100$) to make.
`42 -6x + 7y` customers by the first and `66 + 4x - 7y` buy the second.
Therefore the profit can be written as
`P = (42-6x+7y)(x-20) + (66+4x-7y)(y-20)`
`= (-6x^2 - 162x - 140y + 7xy - 840) `
`+ (-2y^2 + 206y - 80x + 4xy - 1320)`
`= -6x^2 - 2y^2 - 242x + 66y + 11xy - 2160`
To maximise the profit P with respect to x and y we differentiate partially with respect to each, set the derivative to zero and solve.
`(delP)/(delx) = -12x -242 + 11y`
`(delP)/(dely) = -4y + 66 + 11x`
Setting these partial derivatives to zero we get the simultaneous equations
1) `-12x + 11y = 242`
2) `11x - 4y = -66`
Adding 11/12 1) to 2) we get
`(121-48)y = 242(11) - 66(12)`
`implies` `y = 1870/73 = 25.6164 ` x 100$
Substituting into 1)
`12x = 11(1870/73) - 242`
`implies` `x = 3.3151` x 100$
Check: 1) -12(3.3151) + 11(25.6164) = 242, 2) 11(3.3151) - 4(25.6164) = -66
To maximise profit the company should sell the first system for
x = 3.3151 x 100 = 331.51$ and the second for
y = 25.6164 x 100 = 2561.64$
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