Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem12

Video Transcription

12. Consider all rectangles that can be divided into three rectangles, two of which have the dimensions 11 by 7 and 8 by 4.  What should the area of the third rectangle be so that the area of the original rectangle is as large as possible?  So I have drawn here a rectangle measuring 7 by 11 and 8 by 4, arranged them in all the possible arrangements, and then what we will have to do is, in each figure, complete that  into a rectangle by drawing in a rectangle like so in each case, and then computing the area of the original, we'll choose the one that's as large as possible, and then also  compute the area of the smaller rectangle we have added in for our answer. So let's begin with case 1 here.  We have a width of 11 and a height of 11 for an area of 121, then in the second case, we have a width of 8 and a height of 11 plus 4, so 11 times 15 is 120.  Next, we have a height of 11 and a width of 7 plus 8, so 11 times 15 is 165. And finally, we have a length of 11 plus 8 times a width of 7, so 19 times 7 is 133.  So the largest area is 165, but our answer has to be the dimension of the missing rectangle here, and this rectangle has a length of 11 minus 4, so 7 over here, and a width of 8,  just as the red rectangle. So 7 times 8 is the dimension we're looking for, and the answer is D.

Video Summary

Given the problem of dividing a rectangle into three parts with two specific dimensions (11x7 and 8x4), the goal is to determine the area of the third rectangle to maximize the total area. By examining different configurations, the optimal solution yields an original rectangle area of 165 square units. In this arrangement, the third rectangle has dimensions of 7x8, giving it an area of 56 square units. Therefore, the area of the third rectangle, which adds to achieve the maximal total area, is 56.