<p>Thanks for your time guys...having problems with optimization and related ratio problems...</p>
<p>A man has 340 yards of fencing for enclosing two separate fields, one of which is to be a rectangle twice as long as it is wide and the other a square. The square field must contain at least 100 square yards and the rectangle one must contain at least 800 square yards.</p>
<p>a) if x is the width of the rectangular field, what are the maximum and minimum possible values of x?</p>
<p>b) What is the greatest number of square yards that can be enclosed in the two fields? Justify!</p>
<p>first, draw out the rectangle and the square. when i did it, i defined the width of the rectangle as x, both because the problem does so also and for convenience's sake. then it follows that the length of the rectangle is 2x (because it's twice as long as it is wide).</p>
<p>now you have to define the length of the sides of the square in terms of x (in the meantime, though, you can refer to it as y). since the problem tells you that 340 yds of fencing are used, you know that the sum of the perimeters is 340. therefore, you can set up the equation 2(x) + 2(2x) + 4y = 340, where y is the length of the side of the square. solving yields y = 85 - 3x/2.</p>
<p>now that you've defined everything, you can actually optimize the area contained in the fences. first, define the sum of the areas... like so:</p>
<p>A(x) = 2x^2 + (85 - 3x/2)^2</p>
<p>which simplifies to 17x^2/4 - 255x + 7225</p>
<p>then take the derivative...</p>
<p>A'(x) = 17x/2 - 255</p>
<p>this represents the slope of the function defined above, so the maximum area will be located where the derivative equals zero...</p>
<p>solving 17x/2 - 255 = 0 yields x = 30. so the maximum area will occur when the width of the rectangular field is 30. that area is 30(60) + (85 - 45)^2 = 3400 square yards.</p>
<p>a) x is max when the square field is min area. y^2 = 100, so y=10, so perimeter of rectangle = 300, so x = 50. X is min when the area of the rectangle is min. min area is 800. 800 = 2x^2. x = 20.</p>
<p>Thanks a lot guys appreciate the walkthroughs</p>
<p>just one more question and then my homework that finishes my Calc Homework for thanksgiving :) (Had to do 15 of these problems :( )</p>
<p>A tightrope is stretched 30 feet above the ground between the Jay and the Tee buildings, which are 50 feet apart. A tightrope walker, walking at a constant rate of 2 feet per second from point A to point B, is illuminated b a spotlight 70 feet above point A, as shown in the diagram.</p>
<p>a) How fast is the shadow of the tightrope walker's feet moving along the ground when she is midway between the buildings?</p>
<p>b)How far from point a is the tightrope walker when the shadow of her feet reaches the base of the Tee building?</p>
<p>c)How fast is the shadow of the tightrope walker's feet moving up the wall of the Tee building when she is 10 feet from point B?</p>
<p>All i have is the crude drawing below so if you guys can't use this for the problem then i'll understand thanks for trying anyways.</p>
<p>This is an actual former AP problem, named after one of the great AP Calculus question writers.</p>
<p>Honestly, the best thing that I can recommend is to take your diagram and label the different parts more fully. If we call M the place where the spotlight hits the ground (line DC), and if we call Z the location where the spotlight hits the tightrope (line AB), then triangle JAZ and triangle JDM are similar. Keep in mind that parts (a) and (b) really don't require any calculus.</p>
<p>For part (c), keep in mind that there is no intersection for where the spotlight hits the ground, as the spotlight hits the Tee building between points B and C.</p>
<p>This question is a tough problem, but I think these hints will help you along your way.</p>
<p>ok...just a question on the first problem. I'm at the step where you are about to find the derivative. I have 2(x)(x)+ (85-1.5x)^2=A(x) if this is the correct equation then i keep getting 17x/4 not 17x/2...</p>
<p>are you guys sure 17/2 is correct? If so then I'll find anothe rway to do it.</p>
<p>ok so i finally fully understand everything about 1 except for one thing. How do you know that the area of the square is when x is max and the area of the rectangle is when x is min.</p>
<p>It has something to do with how much bang you get for your area buck for the side lengths.</p>
<p>For instance, let's say that you have 3600 m^2 of fencing available for a project:</p>
<p>It's rather easy to see that using a square plot, you could do this by having a plot that is 60m x 60m. This has a perimeter of 240m (60m x 4 sides).</p>
<p>Let's contrast that to a rectangular plot as defined in your question. Here, the sides are 30<em>sqrt(2)m x 60</em>sqrt(2)m (you can find the smallest side by solving for x in 2(x) + 2(2x) = 3600). This has a perimeter of 180sqrt(2), which is approximately 254.558 m.</p>
<p>You can fiddle around with the number for the side length as much as you want, but what you'll find is that as l = w, the perimeter is minimized for a given area, or conversely, that as l = w, the area for a given perimeter is maximized.</p>
<p>In addition, increasing the number of sides of regular polygons will decrease the perimeter of the sides needed. If we let the limit of the number of sides go to infinity, we're really looking at something circular. Using this idea, it should be no surprise that a circle with a radius of approximately 33.851m (this is sqrt(3600/pi)) would produce the smallest "perimeter" of all, where the circumference of the circle yields 212.694m of fencing, a significant improvement over even the square.</p>
<p>Let's go back to your question now:</p>
<p>You have a fixed amount of fencing (perimeter). If you give as much of that fencing as you can to the square, you're going to maximize the total amount of area. Conversely, if you give as much of that fencing as you can to the rectangle, you're going to minimize the total amount of area. Anything in between giving the most possible to one or the other is merely going to give you something in between.</p>
<p>Thank you so much. Yeah, it cleared things up for me. Now not only do i have the answer to this problem but I can explain it to some of my classmates as well :P</p>
