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
Jan 8, 2016

2sqrt222 "in"xx4sqrt2in×42 "in"in

Explanation:

We can write the following equations:

lw=16lw=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)f(l,w)=l2+(w2)2

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

l=16/wl=16w

Thus,

f(w)=sqrt((16/w)^2+(w/2)^2)f(w)=(16w)2+(w2)2

Simplify:

f(w)=sqrt(256/w^2+w^2/4)=sqrt((1024+w^4)/(4w^2))=(sqrt(w^4+1024))/(2w)f(w)=256w2+w24=1024+w44w2=w4+10242w

It should be noted that the domain of this function, or the values for which ww can exist, is 0 < w < oo0<w<.

To find the minimum value, find the derivative of f(w)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.