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 $l = p/4$.