The equation p=4+3y is maximised at (4,5) and the maximised value is 31.
Given equation p=4x+3y and constraints x>0,x<4, y<5.
We have to maximise the equation p=4x+3y with constraints x>0,x<4,y<5.
To find out the points we have to make graphs of the constraints first.
From the graph we have found that points are (0,0),(0,5),(4,0),(4,5).
The above points are common points of graphs of all constraints.
The value of equations are as under:
At (0,0) , p=4*0+3*0=0
At (0,5), p=4*0+3*5=15
At (4,0),p=4*4+3*0=16
At (4,5),p=4*4+3*5=16+15=31
Maximum value of p is 31 and that is for (4,5).
Hence the equation is maximised at (4,5) and the maximum value is 31.
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Answer:
The correct option is D. y = 25.
Step-by-step explanation:
i) From the table and our own observation and a trial and error approach we
can clearly see that the equation that matches y in feet ( the distance
traveled by the rocket) and x in seconds ( the time elapsed) is given by the
equation y = 25
Therefore the correct option is D. y = 25.