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I am doing an maths investigation where I use applied quadratics to find the maximum area that a paddock can be when the perimeter must add to $100$ metres. Using the $A = L \times W$ formula, and writing length in terms of width, a quadratic equation is formed. $W(50-W) = L$ (where $W$ = width and $L$ = length)

Part of the analysis says to discuss the limitations of quadratics in this instance. I have no idea what limitations quadratics have or could have when applied to this problem. (The teacher did mention something about quadratics only being an approximation if that helps.)

Cheers if anyone can shed some light!

Brendan

Hadi
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Breno J
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1 Answers1

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First, your equation needs correction. It should be $A=(50-w)w$. You might like to work through your algebra again to see why.

If you sketch/draw a graph of this equation you should see one way in which the world of the quadratic equation doesn't match with the 'real world' of the problem.

robert timmer-arends
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  • Yes i have sketched up the equation (it was written wrong in the original post) and still can't see how the two don't match – Breno J Feb 29 '20 at 04:01
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    A negative quadratic $A=50(50-w)w$ admits a global maximum (think of a downward-shaped parabola), whereas the opposite is true for the equation you wrote initially. – Hadi Feb 29 '20 at 04:06
  • Yes, that was a typo on my behalf, on my investigation the equation is correct. I still don't understand what the limitations of the quadratic are in this instance? – Breno J Feb 29 '20 at 04:42
  • On the graph, in principle what values can w have (and hence A)? What values can w have for the rectangle? – robert timmer-arends Mar 01 '20 at 06:32