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?

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.