1 / 4

Question 3 D1 January 2012 exam paper

Question 3 D1 January 2012 exam paper. Federico Midolo. In order to do this , we will need to sketch the constraints , represented by the inequailites , on a graph. Let’s start by treating them as normal equations , to find their x and y intercepts.

calvin-foreman
Télécharger la présentation

Question 3 D1 January 2012 exam paper

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Question 3D1 January 2012 exampaper Federico Midolo

  2. In order to do this, wewillneed to sketch the constraints, represented by the inequailites, on a graph. Let’s start by treatingthemasnormalequations, to findtheirx and y intercepts. Note: the signhaschanged to equal, so thatitiseasier to find the x and y intercepts x-intercept y-intercept when x=0, y=11; when y=0, x=11 x+y=11 3x+5y=39 x+6y=39 (11, 0) (13, 0) (39, 0) (0, 11) (0, 7.8) (0, 6.5) when x=0, y=39÷5=7.8;when y=0, x=39÷3=13 Nowwe can plot theselines on a graph(next slide). when x=0, y=39÷6=6.5;when y=0, x=39

  3. y 20 Shade out the unaccaptableregion, to keep the feasibleregionclear and easy to identify. Note: alwayslabel the plottedlines. 15 x-intercept y-intercept Wenowneed to plot the objective line 2x+3y and virtuallymoveitacrossourgraph to find the intersectionpointlocatedfurthestaway (aswe are asked to maximise). x+y=11 (0, 11) (0, 7.8) (0, 6.5) (11, 0) (13, 0) (39, 0) x+y=11 3x+5y=39 x+6y=39 10 This is the maximum point of 2x+3y, when subject to these constraints. We can find its coordinates accurately, by solving the simultaneous equation x+y=11and 3x+5y=39(next page). 3x+5y=39 x+6y=39 5 2x+3y (x and y intercepts) When x=0, y=coefficient of x=2 (0,2) When y=0, x=coefficient of y=3 (3,0) 0 10 20 30 40 x 2x+3y

  4. x+y=11 and 3x+5y=39 x+y=11 y=11-x The answeris x=8, y=3 3x+5(11-x)=39 3x+55-5x=39 2x=16 x=8 y=11-8=3

More Related