Two problems

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• Mar 20th 2010, 12:22 PM
myaa02
Quote:

Originally Posted by icemanfan
That's a start. When you multiply it out, you will find that the function is a parabola. Where does the parabola achieve its maximum value?

by changing $-100x^2-6000x+488000$to f(x) = a(x-h)²+k form ?
• Mar 20th 2010, 12:27 PM
icemanfan
Quote:

Originally Posted by myaa02
by changing $-100x^2-6000x+488000$to f(x) = a(x-h)²+k form ?

Yes, you want to change the equation to that form. Be careful, though. You made a mistake in multiplying it out. The correct equation is $f(x) = -100x^2 + 2000x + 480000$.
• Mar 20th 2010, 12:29 PM
myaa02
Quote:

Originally Posted by icemanfan
Yes, you want to change the equation to that form. Be careful, though. You made a mistake in multiplying it out. The correct equation is $f(x) = -100x^2 + 2000x + 480000$.

how did you get 2000x if you multiply 100x with 80 ?

nvm i just got it lol
• Mar 20th 2010, 12:36 PM
myaa02
what do you do after $-100(x^2-20x+100)+480000-100$ ?
• Mar 20th 2010, 12:40 PM
icemanfan
Quote:

Originally Posted by myaa02
what do you do after $-100(x^2-20x+100)+480000-100$ ?

You should get $f(x) = -100(x^2 - 20x + 100) + 480000 + 10000 = -100(x - 10)^2 + 490000$.

Then, observe that the parabola achieves its maximum value at the vertex. What is the vertex of this parabola?
• Mar 20th 2010, 12:42 PM
myaa02
Quote:

Originally Posted by icemanfan
You should get $f(x) = -100(x^2 - 20x + 100) + 480000 + 10000 = -100(x - 10)^2 + 490000$.

Then, observe that the parabola achieves its maximum value at the vertex. What is the vertex of this parabola?

vertex is -10
• Mar 20th 2010, 12:45 PM
icemanfan
Quote:

Originally Posted by myaa02
vertex is -10

Incorrect. Remember, the vertex is a point. Also, the x-value of the vertex is not -10.
• Mar 20th 2010, 12:48 PM
myaa02
Quote:

Originally Posted by icemanfan
Incorrect. Remember, the vertex is a point. Also, the x-value of the vertex is not -10.

i seriously don't know how to get the vertex.
• Mar 20th 2010, 12:51 PM
icemanfan
Quote:

Originally Posted by myaa02
i seriously don't know how to get the vertex.

If the equation of the parabola is $f(x) = a(x - h)^2 + k$,
then the vertex of the parabola is (h, k).
• Mar 20th 2010, 12:51 PM
myaa02
Quote:

Originally Posted by icemanfan
If the equation of the parabola is $f(x) = a(x - h)^2 + k$,
then the vertex of the parabola is (h, k).

so its (-10,490000)
• Mar 20th 2010, 12:53 PM
icemanfan
Quote:

Originally Posted by myaa02
so its (-10,490000)

The vertex is (10, 490000).
• Mar 20th 2010, 12:54 PM
myaa02
Quote:

Originally Posted by icemanfan
The vertex is (10, 490000).

now how am I supposed to find the " new markup " in order to maximize the revenue ?
• Mar 20th 2010, 12:56 PM
icemanfan
Quote:

Originally Posted by myaa02
now how am I supposed to find the " new markup " in order to maximize the revenue ?

The function is maximized when x = 10. The markup is 100x, so what should the markup be?
• Mar 20th 2010, 12:57 PM
myaa02
Quote:

Originally Posted by icemanfan
The function is maximized when x = 10. The markup is 100x, so what should the markup be?

So, the markup should be 1000 ?
• Mar 20th 2010, 12:59 PM
icemanfan
Quote:

Originally Posted by myaa02
So, the markup should be 1000 ?

Yes, that's correct.
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