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Mathematics 15 Online
OpenStudy (anonymous):

Annual profit in thousands of dollars is given by function, P(x)= -.1x^2 + 50x -300 , where x is the number of items sold

OpenStudy (tkhunny):

Shall we maximize profit? Consider \(\dfrac{-50}{2\cdot (-0.1)}\). Then, tell be where I got that magic number.

OpenStudy (anonymous):

yes, do you want the five values that go with the equation?

OpenStudy (tkhunny):

Do we need them? We have the profit equation and its maximum is well known, at x = 250, as shown above. All that is left is to evaluate P(250).

pooja195 (pooja195):

Are you looking for the steps in which to obtain max profit given quadratic function? If so you can use x = -b/2a where a,b,c correspond to coefficients of function. This gives the x value of where our max occurs, this is where the 2500 units comes from. To get the y value (92500) substitute x value into function. ->P(2500) = 92500 Does that help?

OpenStudy (anonymous):

actually we do because part of the question is find the profit function for the maximum

pooja195 (pooja195):

?

OpenStudy (anonymous):

wait, i'll post as given

OpenStudy (anonymous):

OpenStudy (anonymous):

my answer would depend on the given five values, would it not?

OpenStudy (amistre64):

the 5 values are just a means for you to work the equation ... not for finding its maximum

OpenStudy (amistre64):

without even knowing calculus, what would you determine to be the maximum value of an "upside down" parabola?

OpenStudy (anonymous):

oh! thank you! didn't know that. So, the answer I got was P(250) = 5950. but pooja195 gave To get the y value (92500) substitute x value into function. ->P(2500) = 92500 Which would be correct?

OpenStudy (anonymous):

without knowing calculus, I would say the vertex would be the maximum value

OpenStudy (amistre64):

then determine the value of the vertex :) and that will give you your "x" value that attains the maximum "profit" value

OpenStudy (anonymous):

and how do I determine the value of the vertex? I am sorry to seem so ignorant...but I am.

OpenStudy (amistre64):

do you recall the general setup for a quadratic: ax^2 + bx + c the vertex has an x value of -b/2a; which is what tkhunny alluded to in the first post

OpenStudy (anonymous):

oh! that is where I got the 250

OpenStudy (anonymous):

x = -b/2a = -50/(2* - 0.1) = 250

OpenStudy (anonymous):

Is that right?

OpenStudy (amistre64):

25/.1 = 250 yes so, when x=250, what is our profit? another way to go about this is to do a complete square on it, that reduces it to its vertex form

OpenStudy (amistre64):

y= -.1x^2 + 50x -300 -10y = x^2 -500x +3000 -10y = x^2 -500x +250^2 -250^2 +3000 -10y = (x-250)^2 -250^2 +3000 y = -.1(x-250)^2 +250(25) -300 y = -.1(x-250)^2 + 10(25(25) -30) y = -.1(x-250)^2 + 10(625 -30) y = -.1(x-250)^2 + 10(595) y = -.1(x-250)^2 + 5950 when x=250, y = 5950

OpenStudy (anonymous):

yeah! I was right! Thank you for helping me with this. It is not so much that I don't know what I am doing, but rather having confusion on the component names. I really appreciate your help.

OpenStudy (amistre64):

your welcome, i still have problems with the vernaculars used at times :)

OpenStudy (anonymous):

I find that hard to believe. You seem to know exactly what you are doing. I am a novice in Algebra, but hope one day to be an expert.

OpenStudy (amistre64):

:) keep practicing then

OpenStudy (anonymous):

will do! Thank you again and I hope you have a wonderful day!

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