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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"#

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.

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Answer 2

Ezekiel Alexander

To find the dimensions of the rectangle so that the distance from one corner to the midpoint of a nonadjacent side is minimized, we need to use calculus. Let the dimensions of the rectangle be $x$ and $y$ , with $x$ being the length and $y$ being the width. The area of the rectangle is $A = xy = 16$ . The distance from one corner to the midpoint of the nonadjacent side can be represented by the function $d(x) = \sqrt{x^2 + \left(\frac{y}{2}\right)^2}$ .

To minimize $d(x)$ , we first express it solely in terms of one variable. Since $A = 16$ , we have $y = \frac{16}{x}$ . Substitute this into $d(x)$ to obtain $d(x) = \sqrt{x^2 + \left(\frac{8}{x}\right)^2}$ .

Next, we differentiate $d(x)$ with respect to $x$ and set the derivative equal to zero to find critical points:

\frac{d}{dx} d(x) = \frac{x}{\sqrt{x^2 + \left(\frac{8}{x}\right)^2}} - \frac{8}{x^2}\left(\frac{8}{x}\right) = 0

Solving this equation gives us the critical points. After finding the critical points, we check for local minimums by using the second derivative test.

Finally, we check the boundary points of the feasible domain $0 < x < \infty$ to ensure that our solution is indeed the minimum.

After finding the value of $x$ , we can find the corresponding value of $y$ using $y = \frac{16}{x}$ . These values of $x$ and $y$ will give the dimensions of the rectangle that minimize the distance from one corner to the midpoint of a nonadjacent side.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.