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Calculus/differentiation maxima and minima

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Expert: - 3/13/2010


Question
hi
i understand that when a function reaches a maximum value when it stops increasing and starts decreasing and vice verse for minimum. i am also able to sketch a graph when given y=(x+5)^2(x-1) but now i get stuck at electric potential. i am busy studying through correspondence and are preparing for exam working through some old exam papers and are struggling with the following question and would like to know if you can assist me please (it is not home work)

Q, The electric potential at a point (x;y) on the line segment extending from (0;3) to (2;0) is given by P=3x^2+2y^2.
At what point on this segment is the potential a minimum?
(Hint:the equation of the line segment is y=-3/2x+3)

can you please explain to me where to start and what to look for with such a question

regards
Jacques

Answer
The line segment is y = -3x/2 + 3.

Since P = 3x² + 2y² and the equation is on the line y = -3x/2 + 3, put in that value for y.

This gives P(x) = 3x² + 2(-3x/2 + 3)² = 3x² + 2(9x²/4 - 9x + 9) = 15x²/2 - 18x + 18.

From here, we can see that P'(x) = 15x - 18.

Setting this to 0 to find the extreme points gives 15x = 18, or x = 6/5.

Put that back in the linear equation y = -3x/2 + 3 to find y.

One x, y have been found, both can be put in the original P = 3x² + 2y² to find the value of P.

Note that y need not be found if x is put into P(x) = 15x²/2 - 18x + 18;
or maybe that's a way to check the problem by making sure that same value is gotten for P.

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