# 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?

##### 1 Answer

#### 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

#l=16/w#

Thus,

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

To find the minimum value, find the derivative of

#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

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

The derivative does not exist when

To find the extrema, find the function values for the endpoints of the domain,

Since

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

#f(4sqrt2)=4#

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

Since

Note that since the Pythagorean formula I created used