Annual profit in thousands of dollars is given by function, P(x)= -.1x^2 + 50x -300 , where x is the number of items sold
Shall we maximize profit? Consider \(\dfrac{-50}{2\cdot (-0.1)}\). Then, tell be where I got that magic number.
yes, do you want the five values that go with the equation?
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).
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?
actually we do because part of the question is find the profit function for the maximum
?
wait, i'll post as given
my answer would depend on the given five values, would it not?
the 5 values are just a means for you to work the equation ... not for finding its maximum
without even knowing calculus, what would you determine to be the maximum value of an "upside down" parabola?
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?
without knowing calculus, I would say the vertex would be the maximum value
then determine the value of the vertex :) and that will give you your "x" value that attains the maximum "profit" value
and how do I determine the value of the vertex? I am sorry to seem so ignorant...but I am.
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
oh! that is where I got the 250
x = -b/2a = -50/(2* - 0.1) = 250
Is that right?
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
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
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.
your welcome, i still have problems with the vernaculars used at times :)
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.
:) keep practicing then
will do! Thank you again and I hope you have a wonderful day!
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