Question

A rectangle is to have an area of 16 square inches. How do you find its dimensions so that the distance from one corner to the midpoint of a nonadjacent side is a minimum?

Answer 1

Hint: To solve the above question, we should know about the rectangle. A rectangle is a 2D shape in geometry, having 4 sides and 4 corners. Its two sides meet at right angles. It has 4 angles, each measuring 90 degrees. The sides of a rectangle have the same lengths and are parallel. The area of the rectangle is $(length\times breadth)$ and the perimeter of the rectangle is $2(length+breadth)$.

Complete step-by-step solution:
We have given that the area of the rectangle is $16m^2$.
We can write is also as:
$\begin{array}{rl}& \Rightarrow Area=(length\times breadth)\\ & \Rightarrow 16=length\times breadth\end{array}$
Now by using question we will draw a diagram of the line cutting through the rectangle and use the Pythagorean Theorem which is as:
${\left(Hypotenuse\right)}^2={\left(length\right)}^2\times {\left(breadth\right)}^2$
Let $l$ be the length of the rectangle and $w$ be the breadth of the rectangle

Now we will find the length of hypotenuse say $f(l,w)$, we get
$\Rightarrow f(l,w)=\sqrt{l^2+{\left({\displaystyle \frac{w}{2}}\right)}^2}$
Now by using the area equation $$16=length\times breadth$$we will make $f(l,w)$ into single variable
$\begin{array}{rl}& \Rightarrow 16=(length\times breadth)\\ & \Rightarrow l={\displaystyle \frac{16}{w}}\end{array}$
Now substitute $l$ by ${\displaystyle \frac{16}{w}}$ in $f(l,w)=\sqrt{l^2+{\left({\displaystyle \frac{w}{2}}\right)}^2}$, we get
$\begin{array}{rl}& \Rightarrow f(l,w)=\sqrt{l^2+{\left({\displaystyle \frac{w}{2}}\right)}^2}\\ & \Rightarrow f\left(w\right)=\sqrt{{\left({\displaystyle \frac{16}{w}}\right)}^2+{\left({\displaystyle \frac{w}{2}}\right)}^2}\end{array}$
Now solving the above equation we get
$$\begin{array}{rl}& \Rightarrow f\left(w\right)=\sqrt{{\displaystyle \frac{256}{w^2}}+{\displaystyle \frac{w^2}{4}}}\\ & \Rightarrow f\left(w\right)=\sqrt{{\displaystyle \frac{1024+w^4}{4w^2}}}\\ & \Rightarrow f\left(w\right)={\displaystyle \frac{\sqrt{w^4+1024}}{2w}}\end{array}$$
So by the above equation the value $w$ exists between $0<w<\mathrm{\infty }$.
Since we have to find the minimum value of $w$ so for this find the derivative of $f\left(w\right)$.$$\Rightarrow f\left(w\right)={\displaystyle \frac{\sqrt{w^4+1024}}{2w}}$$
Now by using quotient rule ${\displaystyle \frac{d(u.v)}{dx}}={\displaystyle \frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}}$ on above equation we get
$$\begin{array}{rl}& \Rightarrow f\stackrel{`}{}\left(w\right)={\displaystyle \frac{{\displaystyle \frac{4w^3\left(2w\right)}{2\left(\sqrt{w^4+1024}\right)}}-2\sqrt{w^4+1024}}{4w^2}}\\ & \Rightarrow f\stackrel{`}{}\left(w\right)={\displaystyle \frac{{\displaystyle \frac{4w^4}{\sqrt{w^4+1024}}}-{\displaystyle \frac{2(w^4+1024)}{\sqrt{w^4+1024}}}}{4w^2}}\\ & \Rightarrow f\stackrel{`}{}\left(w\right)={\displaystyle \frac{2w^4-2048}{4w^2\sqrt{w^2+1024}}}\end{array}$$
Now by more simplifying, we get
$$\Rightarrow f\stackrel{`}{}\left(w\right)={\displaystyle \frac{w^4-1024}{2w^2\sqrt{w^4+1024}}}$$
Now set the above equation equals to zero, we get
$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{w^4-1024}{2w^2\sqrt{w^4+1024}}}=0\\ & \Rightarrow w^4-1024=0\\ & \end{array}$
Now add $1024$ on both sides, we get
$\begin{array}{rl}& \Rightarrow w^4=1024\\ & \Rightarrow w=\sqrt[4]{1024}\\ & \Rightarrow w=4\sqrt{2}\end{array}$
Therefore the derivative of $f\left(w\right)$ does not exist when $w=0$.
Now we have to find the extrema, first find the function values for the endpoints of the domain, $0$ and $\mathrm{\infty }$, and for the critical value, $4\sqrt{2}$ .
Since $0$ and $\mathrm{\infty }$cannot putted into $f\left(w\right)$
Therefore, $w=4\sqrt{2}$ is the only critical point on the interval $(0,\mathrm{\infty })$, and it is relatively minimum and it is the smallest breadth value that fits the parameter.

Note: We can go wrong by creating the Pythagorean formula, here I created by using${\displaystyle \frac{w}{2}}$ and $l$, which means that$w=4\sqrt{2}$ is the side being bisected. Also make sure the quotient rule ${\displaystyle \frac{d(u.v)}{dx}}={\displaystyle \frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}}$ which we used is also correct.