Quadratic Functions Word Problem, It's Tricky

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The discussion revolves around a word problem involving a 30 cm wire that is cut to form a square and a rectangle with a 2:1 side ratio, aiming to minimize the total area. Participants explore the algebraic setup, leading to the formulation of area functions for both shapes. The key insight is that the minimum area occurs when the wire is allocated in a specific proportion between the square and rectangle, which can be determined using properties of quadratic functions. The conversation highlights methods for solving the problem, including completing the square and using the quadratic formula to find the minimum area. Ultimately, the focus is on understanding the relationship between wire length, shape dimensions, and area minimization.
  • #31
Mentallic said:
Yes, thankyou eumyang. I hope that's where the confusion was so we can finally settle this :biggrin:

You're welcome! :smile:

@zgozvrm: there is yet another way to find the value -b/2a. You can take the two forms of the quadratic (standard form and vertex form) and set them equal to each other:

ax2 + bx + c = a(x - h)2 + k

Expand the right hand side, and then:
* set the coefficients of the x terms equal to each other, and
* set the constant terms equal to each other,
to solve for h and k. You should get
h = -b/2a and k = c - ah2. (Now note, this is another way to derive the formulas for h and k. We wouldn't be repeating this every time we want to solve a specific problem regarding quadratics.)
 
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  • #32
eumyang said:
Completing the square is not just used for finding the zeros of a quadratic equation. It can also be used to transform a quadratic from standard form:
y = ax2 + bx + c
into vertex form:
y = a(x - h)2 + k.

Once you have the quadratic in vertex form, you can also quickly identify your max/min point.

I get that, but my point is this: once shown that h = -b/2a, there is no longer any need to complete the square for additional problems.

In your example of -4x2 + 16x - 23, we can clearly see that h = -16 / 2(-4) = 2 so why complete the square to get y = -4(x - 2)2 - 7, which shows the same thing? ... it's just extra work.



Besides, in post #9, you stated,
Mentallic said:
For the general quadratic y=ax^2+bx+c, the maximum/minimum is located at x=\frac{-b}{2a}
without mentioning how you got there, until HallsofIvy mentioned "completing the square" and I asked about that method. So, the assumption here, is that you can use this identity without completing the square for quadratics.


Also, the OP didn't state that he was to solve by completing the square until after the topic was brought up.
 
  • #33
zgozvrm said:
... it's just extra work.
The OP said he's already spent 2 hours on this problem. Saving time should be the last of his worries in my opinion.

zgozvrm said:
without mentioning how you got there
I didn't think I would need to because he is studying quadratics and would have stumbled across this result many times by now. On the other hand, you didn't know it so I acted accordingly by showing you how it works.
zgozvrm said:
I use completing the square to find factors of a quadratic.
How does it give you max or min?

Then later down the track the OP said it needs to be solved by completing the square, and you're saying it's just extra work to do so since the result has already been given here.

Obviously the extra work isn't going to go to waste.
 
  • #34
I don't want to get into a huge argument but just wanted to throw in my two cents. The OP's original question is essentially solved already anyway.

The fact that x = -b/2a is the min or max of a quadratic seems to me like one of those useful facts that you pick up when working on math competitions. It's always a good idea to know that it does indeed come out of completing the square (or calculus). And, I do think that is a nontrivial fact in that many people I know (granted, not mathematicians) wouldn't know this off the top of their head.

More important is what is gained from going through the derivations. The reason why completing the square works well in giving a min/max is because you can work the equation into the form (something)^2 + something, and we can use the fact that (something^2) >= 0. In fact, this is in many ways the most basic and important inequality, the starting point of so many other inequalities, so it's nice to see it used.
 

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