Stuck on a test question that I'm not entirely sure how to do. Any help? Ms. Ortiz sells tomatoes wholesale. The function p(x) = -80x^2 + 320x - 10, graphed below, indicates how much profit she makes on a load of tomatoes if she sells them for 4-x per kilogram.
Then it says "what should Ms.Ortiz charge per kilogram of tomatoes to make the maximum profit, and what is the maximum profit she can make on a load of tomoatoes.
My guess is $230 at $3 per kilogram $310 at $2 per kilogram** $320 at $2 per kilogram $320 at $3 per kilogram
min/max values tend to bring about derivatives, or at least a vertex
do you know the structure/shape of a quadratic equation? do you know what a derivative of a function is?
No really no. I've been trying to get help from my teacher but it isn't going too well.
hmm, you should have had some work that relates to finding the roots of a quadratic equation at least. can you tell me some things that you have been studying with regards to quadratic equations?
or, if you just want to try out the options; let x = 2, and then 3 to see what values they produce, then pick the biggest one
I have the formula \[x= \frac{ -b \pm ?\sqrt{(b)^{2}-4(a)(c)}}{2(a)}\]
I got really lost on the parabola and graphs part of the lesson though.
that would tell us an interval, and the middle of that interval would be the x value
\[0=-80x^2 + 320x - 10\] \[0=-8x^2 + 32x - 1\] \[0=8x^2 - 32x + 1\] that might help to reduce the numbers to play with
Ohhh I see! So plug them in to see what higher number you get?
32 +- sqrt(32^2 -4(1)(8)) ----------------------- 2(8) 32 +- sqrt(992) --------------- 2(8) the midpoint (average) of these end points is 32 - sqrt(992) 32 + sqrt(992) -------------- + ------------- 2(8) 2(8) ------------------------------- 2 32+32 --------- 2(2)(8) 2(32) 32 --------- = ---- = 2 2(2)(8) 16
shorrtcut .... yeah, see what x=2, and x=3 get you; then pick the highest value
I got x=3 is 230 and x=2 is 310
So B would be the right answer?
good, then 310 at 2 is correct :)
Thank you so much!
youre welcome, and good luck :)
Join our real-time social learning platform and learn together with your friends!