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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.

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Question 3 D1 January 2012 exam paper

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  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

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