# Calculus Made Easy, Exercise 9.3, maximizing area of rectangle

STATEMENT: A line of length $p$ is to be cut up into $4$ parts and put together as a rectangle. Show that the area of the rectangle will be a maximum if each of its sides is equal to $1/4p$.

PROBLEM: Method #2 down here solves it. My intuition is correct ?

PDF of book is here, page 107.

Method #1: Dividing into 4 equal parts. Area A = $p^2/16$ . Area will be maximum or minimum when derivatives is equal to zero. $$\frac{dA}{dp} = \frac p8 \iff \frac p8 = 0 \iff p = 0$$. That is minimum I guess but we have to find the maximum

Method #2: Since it is a rectangle, area $A$ of rectangle with length $L$ and width $W$, is $L\cdot W$. lets say $L = 0.19p$ and $W = 0.31p$ (because circumference of rectangle is equal to the length of the rope $\iff 2L + 2W = P$ which means $2\cdot 0.19 + 2\cdot 0.31 = 0.38 + 0.62 = 1.0$. Hence our estimation of this length is correct. $$A = 0.19p * 0.31p = 0.0589p^2$$

$$\frac{dA}{dP} = 2 \cdot 0.0589 \cdot p \iff 2 \cdot (fractional) \cdot p \iff 2p \cdot (fraction)$$

Hence $A$ is $2p$ multiplied by some $fraction$ (rational or irrational). $2p$ is kind of constant but this $fraction$ will vary for different pieces of lengths of rope. Hence $A$ will be maximum when $fraction$ will be maximum and maximum value it can have is $1$, which means $L$ and $W$ are equal because:

$$1 = 2L + 2W = 2\cdot \frac P4 + 2\cdot \frac P4 \iff$$. Hence $P/4$ wi give maximum area of rectangle.

In a rectangle, opposite sides must be congruent. Thus, we must cut the line segment of length $p$ into two segments of length $l$ and two segments of length $w$. That is, $$p = 2l + 2w \tag{1}$$ The area of the resulting rectangle is $$A = lw \tag{2}$$ Solving equation 1 for $l$ yields \begin{align*} p & = 2l + 2w\\ p - 2w & = 2l\\ \frac{p}{2} - w & = l \end{align*} Substituting this expression for $l$ in equation 2 allows us to express the area as a function of $w$. $$A(w) = \left(\frac{p}{2} - w\right)w = \frac{p}{2}w - w^2$$

Method 1: We use the First Derivative Test.

Differentiating $A(w)$, while keeping in mind that $p$ is a constant, yields $$A'(w) = \frac{p}{2} - 2w$$ Setting the derivative equal to zero and solving for $w$ yields \begin{align*} A'(w) & = 0\\ \frac{p}{2} - 2w & = 0\\ \frac{p}{2} & = 2w\\ \frac{p}{4} & = w \end{align*} Thus, $w = p/4$ is a critical point.

Since $A'(w)$ is a decreasing function of $w$, the sign of the derivative changes from positive to negative at $w = p/4$. The First Derivative Test tells us that the area function $A(w)$ has a relative maximum at $w = p/4$.

Alternatively, since $p$ is a constant, $$A''(w) = -2$$ so the second derivative is negative at the critical point. By the Second Derivative Test, the area function $A(w)$ has a relative maximum at $w = p/4$.

Notice that $0 < w < p/2$. Since the interval is open, the absolute maximum cannot occur at the endpoints of the interval, so the relative maximum is also an absolute maximum.

When $w = p/4$, $$l = \frac{p}{2} - w = \frac{p}{2} - \frac{p}{4} = \frac{p}{4}$$

Thus, the rectangle with maximum area that can be formed by cutting a line segment of length $p$ into four pieces is the one created by cutting the line segment into four pieces each having length $p/4$.

Method 2: We complete the square.

\begin{align*} A(w) & = \frac{p}{2}w - w^2\\ & = -w^2 + \frac{p}{2}w\\ & = -\left(w^2 - \frac{p}{2}w\right)\\ & = -\left[w^2 - \frac{p}{2}w + \left(\frac{1}{2} \cdot \frac{p}{2}\right)^2\right] + \left(\frac{1}{2} \cdot \frac{p}{2}\right)^2\\ & = -\left[w^2 - \frac{p}{2}w + \left(\frac{p}{4}\right)^2\right] + \left(\frac{p}{4}\right)^2\\ & = -\left(w^2 - \frac{p}{2}w + \frac{p^2}{16}\right) + \frac{p^2}{16}\\ & = -\left(w - \frac{p}{4}\right)^2 + \frac{p^2}{16} \end{align*} This is a quadratic equation with a negative leading coefficient, so the area reaches its maximum value of $p^2/16$ when $w = p/4$. As shown above, if $w = p/4$, then $l = p/4$.

• Your solution is pretty sraightforward. I could not understand the rest because author has not introduced using second derivatives to find the min or max till next chapter – Arnuld Aug 12 '18 at 9:11
• @Arnuld I have added an additional method of solution that does not use calculus. – N. F. Taussig Aug 12 '18 at 9:23
• Thanks for 2nd Method, I get it. The equation is in parabola's Vertex form y = a(x-h)^2 + k. . From what I have experienced so far, Derivatives are more about Algebra and Geometry and less about derivatives themselves – Arnuld Aug 12 '18 at 9:59

The question clearly asks the equivalent:

Among all rectangles of some fixed perimeter $p$, what are the dimensions of the rectangle that has the greatest area?

Your solutions make no sense because you have tried to maximize the area function evaluated for a particular choice of cut. More to the point, you are differentiating with respect to a value that is fixed but unknown, the length $p$ of the line segment, and this is not what is allowed to vary in the question. What is allowed to vary is the length of the first cut for a given length of segment. The result is a "solution" that uses the machinery of differential calculus without actually demonstrating any understanding of what the computations mean or how they relate to finding a solution to the problem.

A proper solution proceeds as follows. Here, we are told that the length of the line segment is some value $p$. For such a segment, there are many ways to partition the segment so as to form a rectangle--some will result in long and narrow rectangles, and some will result in rectangles that are more square-like. Intuitively, we can see that if we select a partition for which one of the two parallel sides is small, then the other pair of sides must be long.

Moreover, a little thought should reveal that the sum of the length and width of any rectangle thus formed from the line segment must always be half the total length of the segment; i.e., if the rectangle has length $l$ and width $w$, then $$l + w = p/2.$$ This is because the total perimeter of the rectangle is $p$, being formed by the entire segment, yet the perimeter in terms of its length and width is $2l + 2w$.

It follows from the above that for a given $p$, say if we choose a positive width $w$ not exceeding $p/2$, the entire rectangle is uniquely determined, for we then have $$l = p/2 - w.$$ The resulting area of such a rectangle, in terms of the width, is the function $$A(w) = lw = (p/2 - w)w.$$ Again, we are not interested in the area as a function of $p$, because we seek to maximize $A$ among all line segments of some fixed length $p$ forming perimeters of rectangles of variable width $w$.

Then by differentiation of $A$ with respect to $w$ and identifying the critical points, we can see when the area attains a maximum: $$0 = \frac{dA}{dw} = p/2 - 2w$$ implies $w = p/4$, thus $l = p/2 - w = p/4$, and the rectangle has equal width and length--it is a square. This completes the proof.

Your second solution, as far as I am able to tell, introduces arbitrary numbers and is as nonsensical as the first.

• I agree with what you said here: -- The result is a "solution" that uses the machinery of differential calculus -- which I could not use because I did not know where to start doing derivatives. I can't agree with your statement that my solution is nonsensical because your solution (before you reached dA/dW) is exactly same as what I did. You used abstracted letters L, W, A while I used concrete numbers. I can't wrap my head around abstracts, I can think very fast when I see something practical, may be because I can't go to shop & ask L cm of rope but I can ask 2.5 m of rope. – Arnuld Aug 12 '18 at 9:04