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

`2sqrt2` `"in"xx4sqrt2` `"in"`

Explanation:

We can write the following equations:

`lw=16`

Draw a diagram of the line cutting through the rectangle and use the Pythagorean Theorem to say that the length of the segment can be found through:

`f(l,w)=sqrt(l^2+(w/2)^2)`

Using the area equation, we can make `f(l,w)` into a single variable equation by substituting.

`l=16/w`

Thus,

`f(w)=sqrt((16/w)^2+(w/2)^2)`

Simplify:

`f(w)=sqrt(256/w^2+w^2/4)=sqrt((1024+w^4)/(4w^2))=(sqrt(w^4+1024))/(2w)`

It should be noted that the domain of this function, or the values for which `w` can exist, is `0 < w < oo`.

To find the minimum value, find the derivative of `f(w)` through the quotient rule (or product rule).

`f'(w)=((4w^3(2w))/(2(sqrt(w^4+1024)))-2sqrt(w^4+1024))/(4w^2)`

`=((4w^4)/sqrt(w^4+1024)-(2(w^4+1024))/sqrt(w^4+1024))/(4w^2)=(2w^4-2048)/(4w^2sqrt(w^2+1024))`

`=(w^4-1024)/(2w^2sqrt(w^4+1024))`

Set the derivative equal to `0`.

`w^4-1048=0=>w=root(4)1024=>w=4sqrt2`

The derivative does not exist when `w=0`.

To find the extrema, find the function values for the endpoints of the domain, `0` and `oo`, and for the critical value(s), `4sqrt2`.

Since `0` and `oo` cannot be plugged into `f(w)`, find the limit as `w` approaches those values.

`lim_(wrarr0)f(w)=oo`

`f(4sqrt2)=4`

`lim_(wrarroo)f(w)=oo`

Since `w=4sqrt2` is the only critical value on the interval, and is a relative minimum, it is also a global minimum and is the smallest width value that fits the parameters. The length is `2sqrt2`, found using the original area formula.

Note that since the Pythagorean formula I created used `w/2` and `l`, which means that `w=4sqrt2` is the side being bisected, which forms a square within the rectangle, which is a common theme in optimization problems of this nature.