Probem16+Genevieve,+Carolyn+C.



a) Show that of all the rectangles with a given area, the one with the smallest perimeter is a square.  Ok, I've figured it out! So we are minimizing the function: P=2l +2w. And our given info is A=lw. Since the problem doesn't actually give a value for A, we can just make up one and say that A=25. So by saying that, we can substitute l in terms of w. 25=lw l=25/w.  So, plug this into the perimeter function and simplify. P=2(25/w) +2w P=50/w +2w  Now, to find it's absolute minimum, let's find the critical values by using the first derivative test. P'= -50/w^2 + 2 0=-50/w^2 + 2 -2=-50/w^2 <span style="color: rgb(80, 0, 199);">-2w^2=50 <span style="color: rgb(80, 0, 199);">w^2=25 <span style="color: rgb(80, 0, 199);">w=5 or <span style="color: rgb(73, 6, 172);"> w=-5 (but since you can't have a negative perimeter or area, w=-25 does not work for this problem) <span style="color: rgb(80, 0, 199);">so w=5 <span style="color: rgb(80, 0, 199);"> <span style="color: rgb(80, 0, 199);">We can prove that this is the absolute minimum by taking the limit of the endpoints of the graph <span style="color: rgb(80, 0, 199);">lim as x--> infinity = infinity and the lim as x-->0 <span style="color: rgb(71, 0, 184);"> infinity so, this means that the graph is concave up, and that at w= 5<span class="Apple-style-span" style="color: rgb(0, 0, 0);"><span style="color: rgb(80, 0, 199);"> there is an absolute minimum. <span style="color: rgb(80, 0, 199);">

<span style="color: rgb(229, 16, 16);">Notice what is happening at each endpoint; at each endpoint we get an end behavior of infinity, but if we plug our critical value (5, for example) into our objective function we get a number that it less than infinity (20, in the case of w = 5). Therefore, since the output at the critical value is less than that at the endpoints, it is our absolute min.

<span style="color: rgb(80, 0, 199);"> <span style="color: rgb(80, 0, 199);">If we plug w=5 into the objective function, we get: <span style="color: rgb(80, 0, 199);">P 50/(5) +2(5) <span style="color: rgb(80, 0, 199);">P=20 <span style="color: rgb(80, 0, 199);"> <span style="color: rgb(80, 0, 199);">and the perimeter of a square is P=4s, and if we say that s=5, we also get P=20. <span style="color: rgb(80, 0, 199);">So, this proves the statement that the rectangle (with a given area) with the smallest perimeter is a square.

Carolyn: If you leave the area as A (which is just some number, not a variable) A=lw l=A/w P=2w+2l P=2w + 2(A/w) P= 2w + 2 (Aw^-1) P'= 2 + 2(-Aw^-2) P'= 2 - 2A/w^2 0=2 - 2A/w^2 -2= -2A/w^2 -2w^2=-2A w^2=A

Area of a square = s^2

b) Show that of all the rectangles with a given perimeter, the one with the greatest area is a square.

In this section, we are trying to maximize the area function. So, our objective function is A=lw, with the given perimeter function P=2l+2w. Like in part a, we don't have a given number for the perimeter, so we can make one up. I said that P=50. So, we can substitute l in terms of w by manipulating the perimeter function. 50=2l+2w 2l=50-2w l=25-w

So, plug this into our area function. A=(25-w)w A=25w-w^2

To find its absolute maximum, we first have to find the critical values, so we use the first derivative test to find the critical values: A'=25-2w 0=25-2w 25=2w w=12.5

To make sure this is an absolute maximum, we can take the limits of the endpoints. lim as x-->0 =0 lim as x-->infinity =negative infinity So, this shows that the graph is concave down, meaning that at w=12.5, there is an absolute maximum.<span style="color: rgb(255, 6, 0);"> <span style="color: rgb(121, 33, 186);">

Now, if we plug this into the Area function for a rectangle, we get: A=25(12.5) -(12.5)^2 A=156.25

The area for a square is A=s^2, and if we say that s=12.5, we get A=156.25, as well!

So, this proves the statement that the rectangle (if given its perimeter) with the greatest area is a square

Carolyn: If you do it without plugging in numbers-- P= 2l + 2w 2l=P-2w l=P/2 - w

A=lw A= (P/2-w)w A=Pw/2 - w^2 A' = P/2 - 2w 0= P/2 - 2w -P/2 = -2w w = P/4

since P for a square = 4w, this makes sense.