the company operates in its factory

-25 X^2 + 7500 x

P=7500x-25x^2

P=25x(300-x)

x=150

P=25*150*150

P is maximum when x is 150

Pick C

selango wrote:-25 X^2 + 7500 x

P=7500x-25x^2

P=25x(300-x)

x=150

P=25*150*150

P is maximum when x is 150

Pick C

Can you please explain, how you came up with 150 ?
Even after this P=25x(300-x), I am unable to visualize which number to try first.

Thanks in advance!

For maxmia /minima conditions we can use second order differentiation methods:

dp/dx = -50x + 7500 ; equate to zero to get value of x

50x = 7500

x = 150

To check x will give max value , try to do second order differentiation of p; it should give d^2P/dx^2 <0



d^2P/dx^2 = -50 which is less than zero

pick X= 150

gmatmachoman wrote:For maxmia /minima conditions we can use second order differentiation methods:

dp/dx = -50x + 7500 ; equate to zero to get value of x

50x = 7500

x = 150

To check x will give max value , try to do second order differentiation of p; it should give d^2P/dx^2 <0



d^2P/dx^2 = -50 which is less than zero

pick X= 150

Wow!! Its been more than 10 yrs, I forgot derivatives.. is derivative of x^2 , 2x?
And is there any simpler method? I am sure GMAT doesnt want to test us on differential equations. Excellent approach though, no offence on your approach.

Thanks!

In that case , u need to start plug in numbers which is really time consuming...

Since i work with derivatives day in day out....its a part of me these days..LOL

thirst4edu wrote:

gmatmachoman wrote:For maxmia /minima conditions we can use second order differentiation methods:

dp/dx = -50x + 7500 ; equate to zero to get value of x

50x = 7500

x = 150

To check x will give max value , try to do second order differentiation of p; it should give d^2P/dx^2 <0



d^2P/dx^2 = -50 which is less than zero

pick X= 150

Wow!! Its been more than 10 yrs, I forgot derivatives.. is derivative of x^2 , 2x?
And is there any simpler method? I am sure GMAT doesnt want to test us on differential equations. Excellent approach though, no offence on your approach.

Thanks!

Since P = -25 X^2 + 7500 x = 25 x (300 - x); and in order to maximise P, we need to maximise the product of the two unknowns x and 300 - x, which is maximum when x = 300 - x or at [spoiler]x = 150[/spoiler], P is maximum.

Learn this very important property of two positive numbers a and b, that their product is maximum when a = b.

[spoiler]C[/spoiler]

Differential Calculus techniques are not tested on GMAT, Govind! Please do not offer minors, your toys.

sanju09 wrote: Learn this very important property of two positive numbers a and b, that their product is maximum when a = b.

Thanks Sanju!! Yesterday, I did come close to equation 25x(300-x), and thought there has to be something to do with balancing between x & (300 - x).. but was not sure. Your statement explains it all!

Thank You!!

thirst4edu wrote:

sanju09 wrote: Learn this very important property of two positive numbers a and b, that their product is maximum when a = b.

Thanks Sanju!! Yesterday, I did come close to equation 25x(300-x), and thought there has to be something to do with balancing between x & (300 - x).. but was not sure. Your statement explains it all!

Thank You!!

Thanks thirst4edu, actually my definition about the rule is not in its full, it in fact says that if the sum of two positive numbers is constant, their product is maximum when the numbers are equal.