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...
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 $2000 per system, how should the company price the systems to generate the largest possible profit? Round your answers to the nearest cent.
The price of the first system is `x` x 100$ 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$