{"id":273,"date":"2019-09-20T12:43:27","date_gmt":"2019-09-20T17:43:27","guid":{"rendered":"http:\/\/blogs.acu.edu\/mu_sigma\/?p=273"},"modified":"2019-09-27T11:13:28","modified_gmt":"2019-09-27T16:13:28","slug":"problem-3-fall-2019-maximize-the-area","status":"publish","type":"post","link":"https:\/\/blogs.acu.edu\/mu_sigma\/2019\/09\/20\/problem-3-fall-2019-maximize-the-area\/","title":{"rendered":"Problem 3 – Fall 2019 – Maximize the Area"},"content":{"rendered":"

Maximize the Area<\/b><\/span><\/p>\n

This problem is from the newsletter for users of the TI graphing calculators.<\/i><\/span><\/p>\n

Given a rectangular piece of paper, label the four corners of the paper A<\/b>, B<\/b>, C<\/b>, and D<\/b> (see figure below). Fold the vertex A<\/b> so that it just touches the segment CD<\/b>. Determine the location of point E<\/b> that maximizes the area of the triangle EDF<\/b>.<\/p>\n

<\/p>\n

Please submit all work for this problem to mathpotw@acu.edu no later than 5:00 PM on Thursday, September 26th.<\/p>\n

Solution to Maximize the Area<\/b><\/span><\/p>\n

Correct solutions were submitted by: Bethany Witemeyer.<\/i><\/span><\/p>\n

We let the distance from A<\/b> to D<\/b> be 1 and the distance from F<\/b> to D<\/b> be denoted by x<\/i>. It follows that the distance from F<\/b> to E<\/b> is 1-x<\/i>, since it is equal to the distance from A<\/b> to F<\/b>. Therefore we have a right triangle with height x<\/i> and hypotenuse of length 1-x<\/i>. Thus the square of the length of the base is (1-x<\/i>)x<\/i> – x<\/i>2<\/sup>= 1-2x<\/i>. We want to find x<\/i> that maximizes the area of this triangle.<\/p>\n

It will be easier if we maximize the square of the area since we then have no square roots to differentiate. Thus we differentiate AREA2<\/sup> = x<\/i>2<\/sup>(1-2x<\/i>)\/4 with respect to x<\/i> and set the result equal to zero. Since x<\/i> = 0 would give us a “triangle” with area zero, we know that x<\/i> is not zero. So the solution is x<\/i> = 1\/3. Therefore the distance from D<\/b> to E<\/b> is SQRT(3)\/3.<\/p>\n

Another solution:<\/b> A momentary contemplation allows one to observe that the triangle of maximum area is a 30-60-90-degree triangle. It makes one conjecture that there is a geometric proof not using calculus. Indeed, Alexander Karabegov submitted such a solution. He merely reflected the triangleDFE<\/b> through the segment DE<\/b> to obtain an isosceles triangle with sides of length 2x<\/i>, 1-x<\/i>, and 1-x<\/i>. The sum of the three sides is 2 (independent of x<\/i>), so we see that we are attempting to find an isoceles triangle of maximum area with fixed perimeter 2. Some persons know that such a triangle is an equilateral triangle, hence the same result as before.<\/p>\n","protected":false},"excerpt":{"rendered":"

Maximize the Area This problem is from the newsletter for users of the TI graphing calculators. Given a rectangular piece of paper, label the four corners of the paper A, B, C, and D (see figure below). Fold the vertex A so that it just touches the segment CD. Determine the location of point E […]<\/p>\n","protected":false},"author":130,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[181534],"tags":[],"class_list":["post-273","post","type-post","status-publish","format-standard","hentry","category-problem-of-the-week"],"_links":{"self":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/273","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/users\/130"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/comments?post=273"}],"version-history":[{"count":2,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/273\/revisions"}],"predecessor-version":[{"id":286,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/273\/revisions\/286"}],"wp:attachment":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/media?parent=273"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/categories?post=273"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/tags?post=273"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}