# Intuitive understanding of maximum value of quadratic function

In trying to understand why the maximum area of a rectangle with a fixed perimeter occurs when the base is equal to the height, I got as far as this expression:

$A = (p/2)x - x^2$

from

$p = 2x + 2y,x + y = p/2, y = p/2 -x, A = x(p/2 - x)$

I know that I need to find the maximum value of the top expression, which can be generalised to $bx - x^2$

My question is, is there an intuitive (possibly visual) way to understand how to find the greatest possible value of this expression? Ideally I'd like to avoid anything but the most basic algebraic steps, and not have to refer to graphs or the quadratic equation.

• Have you heard of the arithmetic-geometric mean inequality?
– πr8
May 2, 2017 at 9:21

Assume you have a rectangle with perimeter $p$ such that its sides measure $L$ and $l$ and $L > l$. Let $$A = L \, l$$ be the area.

Now let's modify a bit the lengths of the sides. We want to decrease a bit the big one and to increase a bit the small one, but not too much. Let's see how the area change. Choose $\epsilon > 0$ such that $L- l > \epsilon$. Consider a new rectangle of sides $L-\epsilon$ and $l + \epsilon$. Note that the perimeter is still the same! The new area $A'$ is: $$A' = (L-\epsilon)(l + \epsilon) = A + \epsilon(L-l - \epsilon) > A.$$ This explain intuitively why making the two lengths more similar, the area is increasing.

A quadractic equation $$q(x) = a x^2 + bx + c$$ with non-zero coefficient $a$ has either a minimum or a maximum, depending on the sign of $a$.

If you plot the graph you will see the typical parabola shape.

One way to determine the extremum is to bring $q$ into the form $$q(x) = a (x - S)^2 + T$$ where $(S, T)$ are the coordinates of the extremal point (vertex).

In your case we have \begin{align} A(x) &= -x^2 + (p/2) x \\ &= -(x^2 - (p/2) x) \\ &= -((x - p/4)^2 - (p/4)^2) \\ &= -(x- p/4)^2 + (p/4)^2 \end{align} from which you can read the coordinates of the extremum.

The rate at which the area is changing $dA/dx$ is $p/2 - 2x$ and has a critical value at either $0$ or when $x$ is $1/4p$ so if $2 x$ values are used to preserve the parallels of the rectangle the remaining $p/2$ is the $y$ values which are just $x$ again.

Have a look at the figure below.

The old area is $b \times h$.

The new area is

$$\tag{1}(b-2d) \times (h+2d) =\begin{cases} bh -2 dh + 2d(b-2d)\\bd + 2(b-h)d-4d^2 \end{cases}.$$

In the first form, we recognize $-2hd$ as the (negative!) area of the two "lost" lateral rectangles, and $2d(b-2d)$ which is the area of the two "gained" rectangles).

Let us consider the second formula of (1). Neglecting the "very small quantity" $4d^2$ ($d>0$ is considered as already small), if quantity $b-h$ is positive, the new area is bigger than the old one (one adds a positive number) whereas, if it is negative, the new area is smaller.

Thus the "equilibrium point" is reached for $b=h$, i.e., a square.

Remark: what I just wrote is in the spirit of the "founding fathers" of calculus (even before Leibniz and Newton, i.e., Huyghens, Pascal, Fermat...).

Let's place the rectangle with the sides being parallel to the axes and the origin in the middle, and look at top-right quarter of the rectangle. The condition that the perimeter $p$ is fixed meas that the top-right corner is on the line $x+y=p/4$. If we move the corner along the line then the area change is proportional to $Y-X$, and the maximal area is attained when $X=Y$.