Could someone briefly explain step number four, i am currently confused as to how that steps goes on the teaching sheet. it would mean a lot, thanks.

Step 4: Find the derivative with respect to only ONE variable Since 200cm = x * y , we can rearrange to get 200/x = y, so we can sub this for y Therefore Aofposter = (x + 8) * (200/x + 16) = 200 + 1600/x + 16x + 128 or 16x + 1600x"1 + 328 Differentiating with respect to x gives: A' = 16 - 1600x-2 , or 16 - 1600/x2 Letting A' = 0 and solving for x (the critical points) gives: 0 = 16 - 1600/x2 x2 = 100 Remember the domain of x >= 0: the -16 = -1600/x2 length cannot be a negative number, x = + - 10 therefore we only account for the positive -16x2 = -1600 root in this situation, so x = 10 Step 5: Solve for length and width and find out the maximum and minimum points. Since length of the large rectangle = x + 8 and x = 10 x + 8 = 10 + 8 = 18 Finding if points are min or max by Therefore, the length is 18cm. subbing a number between 0 and 10 and 10 and 20 into a' Finding width: Subbing y back into the smaller rectangle area: 200/x = y O 200/(10) = y 10 20 y = 20 After subbing in test numbers in d' Since width of the large rectangle = y + 16 and y = between 0 and 10 (which got a 20 negative number) and 10 and 20 (which y + 16 = 20 + 16 = 36 got a positive number), it was determined that x is a minimum point Therefore, the width is 36cm. and therefore x + 8 is a minimum measurement. Therefore, the dimensions of the poster of minimum area are 36cm by 18cm.