find 2 numbers whose sum is 12 and whose product is the maximum possible value. (hint: let x be one number, then 12-x the other number. Form a quadratic function by multiplying them and then find the maximum value of the function
Can you use a graphing calculator, or is this by hand?
If you're doing this by hand, as the hint states, create a quadratic, which is y=x(12-x).
does not specifically say, i guess i can use both, but don't know how
Ok.
Did you learn Calculus 1?
no, i jumped from math 095 to math 103, all this is new to me... thanks for the help
Ok. Then, let's do it the graphing calculator way.
Isn't the answer just 6 for both numbers?
Yes...
both numbers
Basically, x=6
Do you know how, though?
well, actually, i'm trying to figure it out and got 6 and -6 not sure if that's right... how did you get just 6
Do you have a graphing calculator right now?
oh wait, did you just pick 6 because it asks for the MAXIMUM value of the function
yes i do
Yes to your first question. If you want to verify it, follow the steps. 1. Graph y=x(12-x) 2. Set your window as the following; domain=from 0 to 12, range=from -10 to 40 3. You should see a vertex, which is your maximum point. 4. Press 2nd+Trace, and press maximum. 5. Set your lower bound and upper bound, so that the vertex is in between. 6. The x-coordinate of the vertex is your x-value that makes the biggest possible product.
That is if you have ti-83 or ti-84
yes ti-84
Then, follow the steps to get x=6.
sorry, don't know how to figure it out on calculator either
\[ y=x(12-x)= -x^2+12 x = -x^2 +12 x -36 +36= -(x-6)^2+36 \] This a parabola of vertex (6,36} turning down. The maximum is when both numbers are equal to 6. 6+6 =12 6 x 6 =36
where did you get the 36?
just to complete the square.
I added 36 and subtracted 36 so I added 0
Have you studied parabolas?
yes i have, a little confused about it, i but i think i got it
Good
thank you
YW
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