**Rectangle of maximum area**

Let’s say that we have a rectangle with a width

of \(x\) and a height of \(y\). By varying \(x\) and \(y\), we can generate various different rectangles. But let’s say that we’re only interested in the particular set of rectangles where all of those rectangles have the same perimeter \(p\). If \(2x+2y\) is the perimeter of any one of these rectangles, then \(2x+2y\) must always add up to \(p\). Thus,

$$2x+2y=p,\tag{1}$$

For any rectangle in the set. A long time ago, the mathematician Pierre de Fermat considered the following problem: if \(xy\) is a set of rectangles such that \(2x+2y=p\), which type of rectangle \(xy\) has the minimum area \(A\). It turns out that the answer to this question is simply just a square where \(x=y\). But let’s use the techniques of finding the minima and maxima of a function to prove this. Since the techniques we developed in previous lessons for finding the minima and maxima of a quantity involve taking the derivatives of some *single-variable* function \(f(x\), the first thing that pops into my mind that we should do to try to find the maximum \(A\) is to express \(A\) in terms of only one variable. Let’s algebraically manipulate Equation (1) to solve for \(y(x)\) (and then, subsequently, plug \(y(x)\) into the equation \(A=xy\)) in order to accomplish this. Using Equation (1), we can express \(y\) as

$$y=\frac{p-2x}{2}.\tag{2}$$

If we plug Equation (2) into (1), then we’ll have

$$A=x\biggl(\frac{p-2x}{2}\biggr)$$

or

$$A=\frac{p}{2}x=x^2.\tag{3}$$

If we take the derivative with respect to \(x\) on both sides of Equation (3), we’ll have

$$A’(x)=\frac{p}{2}-2x.\tag{4}$$

Setting \(A’(x)=0\), we have

$$\frac{p}{2}-2x=0.\tag{5}$$

The value of \(x\) associated with the minimum *or *maximum value of \(A\) is the one which satisfies Equation (5). But to verify whether or not the \(x\) in Equation (5) is associated with a maximum value of \(A\) instead of a minimum, or vice versa, let’s take the derivative on both sides of Equation (5) to see whether \(A’’(0)\) is positive or negative. Doing so, we have

$$A’’(0)=-2.\tag{6}$$

Since \(A’’(0)<0\), it follows that the value of \(x\) associated with Equation (5) must be the \(x\)-value for which \(A\) is a maximum. Solving for \(x\) in Equation (5), we have

$$x=\frac{p}{4}.\tag{7}$$

If we substitute Equation (7) into Equation (1), we have

$$2\biggl(\frac{p}{4}\biggr)+2y=p.$$

Using algebra to solve for \(y\), we have

$$2y=p-\frac{p}{2}=\frac{p}{2}$$

and

$$y=\frac{p}{4}.$$

Thus, the particular type of rectangle with the máximum área such that \(2x+2y=p\) is a square with a width of \(x=\frac{p}{4}\) and a height of \(\frac{p}{4}\).

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