QUESTION 3 D1 JANUARY 2012 EXAM PAPER Federico Midolo.

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Presentation transcript:

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. x+y = 11 3x+5y = 39 x+6y = 39 when x=0, y=11; when y=0, x=11 when x=0, y=39÷5=7.8; when y=0, x=39÷3=13 when x=0, y=39÷6=6.5; when y=0, x=39 x-intercept y-intercept (11, 0) (13, 0) (39, 0) (0, 11) (0, 7.8) (0, 6.5) Note: the sign has changed to equal, so that it is easier to find the x and y intercepts Now we can plot these lines on a graph (next slide).

x-intercept y-intercept (11, 0) (13, 0) (39, 0) (0, 11) (0, 7.8) (0, 6.5) x+y = 11 3x+5y=39 x+6y= x y x+y=11 3x+5y=39 x+6y=39 Note: always label the plotted lines. Shade out the unaccaptable region, to keep the feasible region clear and easy to identify. We now need to plot the objective line 2x+3y and virtually move it across our graph to find the intersection point located furthest away (as we are asked to maximise ). 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) 2x+3y 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=11 and 3x+5y=39 (next page).

x+y=11 and 3x+5y=39 x+y=11 y=11-x 3x+5(11-x)=39 3x+55-5x=39 2x=16 x=8 y=11-8=3 The answer is x=8, y=3