Explain word problem involving quadratics

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Solution 1

Your function tells you the area of your rectangular garden if you use all 200 meters of fencing and it has width w.

This function can be obtained by manipulating the relationship between length, width, the perimeter, and the area.

Let $l$ be the length and $w$ be the width. Then $$2(l+w)=200,$$ since your perimeter is the sum of the lengths of all four sides, and $$lw=Area.$$

Now, we manipulate the perimeter equation to get \begin{align}l+w&=\frac{200}{2}\\ l&=200/2-w\\ &=100-w. \end{align}

Substituting this into the area expression gives $$Area=(100-w)w=100w-w^2.$$

Solution 2

What does it mean to have the area of a rectangle being modeled by a function of its width?

Let the garden's width be $w$, clearly we must have $0<w<100$. Then the other side of it is doomed to be $100-w$. So the garden's area $A$ is given -- or "modeled" -- by the function $A(w)=w(100-w)$.

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Updated on August 01, 2022

Comments

  • Max
    Max over 1 year

    A word problem from Khan academy reads like this:

    Marquise has $200$ meters of fencing to build a rectangular garden.

    The garden's area (in square meters) as a function of the garden's width $w$ (in meters) is modeled by:

    $A(w) = −w^2​​ +100w$

    What is the maximum area possible?

    Since the problem talks about maximum and the equation is a quadratic, I factored the thing to vertex form:

    $$y = -1(w - 50)^2 + 2500$$

    And the solution they are looking for is indeed $A(50) = 2500$ (square meters).

    But I don't really understand what is modeled here. What does it mean to have the area of a rectangle being modeled by a function of its width? Somehow I'm stumped by this example.

  • Max
    Max almost 6 years
    I think I get it now, the function tells me if I make one side $w$-meters long, what is the resulting area, given that I have at most 200 m fence in total. Now we find the maximum for $w$, that is, when all sides are equally long.