I have added the options as well. Chegg has let me down. Please help me understand how to do optimization problems like these! Thank you Bartleby! Choose 1 answer:(А 1.25 х6.75(в2 x 62.5 х 5.5D 3 x 5There is no such rectangle. Given right AABC with AC = 8 and BC = 10:A rectangle is constructed in AABC as shownin the diagram. What are the dimensions of the rectangle that would minimize the rectangle's perimeter?BAC

Name: I have added the options as well. Chegg has let me down. Please help m
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: I have added the options as well. Chegg has let me down. Please help me understand how to do optimization problems like these! Thank you Bartleby! Choose 1 answer:(А 1.25 х6.75(в2 x 62.5 х 5.5D 3 x 5There is no such rectangle. Given right AABC with A

In the figure shown as - ABC is the triangle with,AC = 8 cmBC = 10 cmLet,Length of the rectangle = x Breadth of the rectangle = yFrom the figure we can see that, BE = BC-EC = 10-y -------------------------------------------------------- Now, In ∆ABC and ∆BDE,These two triangles are similar to each other.When two triangles are similar, then their sides are proportional to each other. then,BCBE = ACFC1010-y = 8x10x = 810-y10x8 = 10-yy = 10-5x4 .........................(1) ------------------------------------------------------------------ Perimeter of the rectangle, P = 2Length+BreadthP = 2x+y = 2x+2yFrom equation (1) -P = 2x+210-5x4 = 2x+20-52xP = 20-12x ------------------------------------------------------------------ Differentiating the perimeter P with respect to x, to minimize the rectangle's perimeter.dPdx = ddx20-12xdPdx = 200-121x1-1 Use ddxAxn = Anxn-1dPdx = -12 Here we see that the derivative of perimeter, does not depend on any variable it's a constant.With increase in the value of 'x', the perimeter will keep on decreasing. When, x = 8 (Maximum possible value), P = 16 (minimum)But y = 0 (So no width of the rectangle).Hence, the no rectangle possible. Concept - Here,Area of rectangle P = 2x+yTo minimise any perimeter, first job is to get the expression in one variable.Here, Perimeter has two variables x and y, so we completely converted it into x variable.To maximize or minimize, we always have to differentiate the subject parameter, with respect to any of the desired variable.Here, I differentiated with respect to x. One can convert expression in y variable, and then can differentiate with respect to y.Then, put the derivative equals to zero, then get a value of the variable. Using that value, one can evaluate all the required value of the parameter.In case the derivative is independent of any variable, then that figure or shape does not exists.