Athena 21,718 Posted December 16, 2015 Share Posted December 16, 2015 These are optimization/related rates problems and I really need help. Link to comment Share on other sites More sharing options...
Bearclaw 3,543 Posted December 16, 2015 Share Posted December 16, 2015 http://www.sosmath.com/CBB/viewtopic.php?t=3199 lol Link to comment Share on other sites More sharing options...
Wasta 5,874 Posted December 16, 2015 Share Posted December 16, 2015 I can help you with the first problem. Do I get half the won? Link to comment Share on other sites More sharing options...
Aearom 1,071 Posted December 16, 2015 Share Posted December 16, 2015 I think the first one answer is 32 I am not sure though since it has been ages. Area = 2xy Substiture y Area = 24x-2x^3 take derivative wrt x and equate with zero. 24 - 6x^2 x = 2 Area = 2*2*8 = 32 EDIT: SO embarrased. Don't know why I took 2y too. Link to comment Share on other sites More sharing options...
Athena 21,718 Posted December 16, 2015 Author Share Posted December 16, 2015 I can help you with the first problem. Do I get half the won?Yes (or more maybe). But I want it explained pls. Link to comment Share on other sites More sharing options...
Wasta 5,874 Posted December 16, 2015 Share Posted December 16, 2015 Yes (or more maybe). But I want it explained pls. Area of rectangle first quadrant A = lh A = x(-x^2 + 12) A = -x^3 + 12x A' = -3x^2 + 12 0 = -3x^2 + 12 3x^2 = 12 x = 2 A = -(2)^3 + 12(2) (plugging back into original area equation) A = -8 + 24 A = 16 Double this area to get the full area of the rectangle Max area = 32 Link to comment Share on other sites More sharing options...
chemicalx 9 Posted December 16, 2015 Share Posted December 16, 2015 Is that Calculus AB or BC? Link to comment Share on other sites More sharing options...
Wasta 5,874 Posted December 16, 2015 Share Posted December 16, 2015 I think the first one answer is 64 I am not sure though since it has been ages. Area = 2xy Substiture y Area = 24x-2x^3 take derivative wrt x and equate with zero. 24 - 6x^2 x = 2 Area = 2*2*8 = 32 EDIT: SO embarrased. Don't know why I took 2y too. Ayyy you know I answered it first. Link to comment Share on other sites More sharing options...
Aearom 1,071 Posted December 16, 2015 Share Posted December 16, 2015 Ayyy you know I answered it first. It is okay. When I started writing your answer was not there. You had just asked will you get half won. So... Anyway I answered it wrong so I am just curling up in shame right now. Link to comment Share on other sites More sharing options...
Ddeona Oneul Bame 61,630 Posted December 16, 2015 Share Posted December 16, 2015 I used to know how to do this~ Now it's just Link to comment Share on other sites More sharing options...
Athena 21,718 Posted December 16, 2015 Author Share Posted December 16, 2015 mess. I was overthinking the first one. I started doing A = 2xy but got confused on how to incorporate the quadratic. EDIT: Are there boundaries for the first problem? Like, we're supposed to figure that out before doing the derivative. ex. 0 < x < ? Thanks guys. I'll give you all 300k won (you too bearclaw) Link to comment Share on other sites More sharing options...
Aearom 1,071 Posted December 16, 2015 Share Posted December 16, 2015 mess. I was overthinking the first one. I started doing A = 2xy but got confused on how to incorporate the quadratic. Thanks guys. I'll give you all 300k won (you too bearclaw) Me too? You are so generous. If you need any calculus help ask me. I will not be as hasty from now on and I have a lot of time on hand currently Link to comment Share on other sites More sharing options...
Wasta 5,874 Posted December 16, 2015 Share Posted December 16, 2015 mess. I was overthinking the first one. I started doing A = 2xy but got confused on how to incorporate the quadratic. EDIT: Are there boundaries for the first problem? Like, we're supposed to figure that out before doing the derivative. ex. 0 < x < ? Thanks guys. I'll give you all 300k won (you too bearclaw) the boundaries are where the parabola hits the x axis. So when y = 12 -x^2 0 = 12 - x^2 x^2 = 12 solve for x (negative and positive are your boundaries.) Link to comment Share on other sites More sharing options...
seoIhyun 11,629 Posted December 16, 2015 Share Posted December 16, 2015 Holy crap I am learning optimization currently as well. Yeah fuck that, thank god my teacher is giving us a take home quiz Link to comment Share on other sites More sharing options...
Aearom 1,071 Posted December 16, 2015 Share Posted December 16, 2015 mess. I was overthinking the first one. I started doing A = 2xy but got confused on how to incorporate the quadratic. EDIT: Are there boundaries for the first problem? Like, we're supposed to figure that out before doing the derivative. ex. 0 < x < ? x should be less that square root of 12 (modulus of) I think? the boundaries are where the parabola hits the x axis. So when y = 12 -x^2 0 = 12 - x^2 x^2 = 12 solve for x (negative and positive are your boundaries.) You are so fast I hate you Link to comment Share on other sites More sharing options...
Wasta 5,874 Posted December 16, 2015 Share Posted December 16, 2015 x should be less that square root of 12 (modulus of) I think? You are so fast I hate you Lol! And no problem earlier, we all make mistakes. Link to comment Share on other sites More sharing options...
Aearom 1,071 Posted December 16, 2015 Share Posted December 16, 2015 Lol! And no problem earlier, we all make mistakes. But I always make these silly mistakes. Waste of marks because of lack of concentration tbqh Link to comment Share on other sites More sharing options...
Athena 21,718 Posted December 17, 2015 Author Share Posted December 17, 2015 Donated! Thanks guys~ Link to comment Share on other sites More sharing options...
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