1. ## Problem understanding question

I have a problem:

1. Find two positive numbers whose sum is 300 and whose product is a maximum.

What I don't understand is that last constraint: ''is a maximum''. What do they mean by a product being a maximum? Thanks.

2. ## Re: Problem understanding question

We have x + y = 300 and want to maximize x*y. From the first condition we get y = 300 - x. Now substitute.

x(300 - x) = -x^2 + 300x

This is a parabola that opens DOWNWARD therefore the maximum occurs at the vertex. The x co-ordinate of the vertex is given by

$\displaystyle x \ = \ -\frac{b}{2a} \ = \ -\frac{300}{-2}$

Can you continue?

3. ## Re: Problem understanding question

Ah, I understand now what they mean.

Can't I simply take the derivative of the substitution and set it to 0 to come to a conclusion? (That's what the class is all about at the moment).

As so:

$\displaystyle \frac{d}{dx}-x^2+300x=-2x+300\\\\-2x+300=0\\\\x=150$

Which would imply that the largest number is given by

$\displaystyle 150\cdot(300-150)=\\\\150\cdot150$

? (btw can someone tell me why the first part of these spaces always has an indent?)

Just one more question if this all checks out: Don't I have some specific range as well? I can't seem to put my finger on that range.

4. ## Re: Problem understanding question

Yes you can use derivative, wasn't sure if this was a calculus or pre-calculus question. The positive values for both x and y occur between the 2 roots of the parabola and nowhere else as a quick sketch will show so,

$\displaystyle 0 \ < \ x \ < \ 300$

and

$\displaystyle 0 \ < \ y \ < \ 300$

That range for y is a bit sneaky... if you pick x to range between 0 and 300 then y will range between 0 and 150 but i suppose the roles of x and y can be reversed without fear so technically you get both between 0 and 300.

The indent may have something to do with your settings but i don't know what, i'm using the same code as you but i don't get that indent.

P.S. If you're going to use calculus you will need to prove x = 150 gives the maximum by using the 1st derivative test.

5. ## Re: Problem understanding question

Originally Posted by agentmulder
Yes you can use derivative, wasn't sure if this was a calculus or pre-calculus question. The positive values for both x and y occur between the 2 roots of the parabola and nowhere else as a quick sketch will show so,

$\displaystyle 0 \ < \ x \ < \ 300$

and

$\displaystyle 0 \ < \ y \ < \ 300$

That range for y is a bit sneaky... if you pick x to range between 0 and 300 then y will range between 0 and 150 but i suppose the roles of x and y can be reversed without fear so technically you get both between 0 and 300.

The indent may have something to do with your settings but i don't know what, i'm using the same code as you but i don't get that indent.

P.S. If you're going to use calculus you will need to prove x = 150 gives the maximum by using the 1st derivative test.
By 'first derivative test' do you mean this?

$\displaystyle f'(149)>0 , f'(151)<0$ Therefore x=relative maxima.

?

6. ## Re: Problem understanding question

Yes, you are correct.

7. ## Re: Problem understanding question

Originally Posted by agentmulder
Yes, you are correct.

Thank you very much!