Maxima and minima in calculus
Learn the concept of maxima and minima using differential calculus with clear rules, first and second derivative tests, and solved examples. Essential for NEET, JEE Main, and JEE Advanced exam preparation.
Maxima and minima concept:
Let us consider, a physical quantity y depends on another physical quantity x i.e. y = ƒ(x). A graph shown in figure represents the variation of y with respect to x.
From the graph, point A represents the maximum value of y when x = x1 and point B represents the minimum value of y when x = x2. Now at point A and B we draw two tangents and the slope of the tangents at those points are tan
= 0.
We know that tan = 
.
Therefore, for maximum and minimum value of y, slope of the tangents = tan = = 0.
Now we consider points C and D just before and after point A respectively.
At point C slope is positive but at point D slope is negative. Therefore, slope decreases at point A with respect to change of x.
Therefore, for maximum value of y, 
= 
= 
< 0.
Now we consider points E and F just before and after point B respectively. At point E slope is negative but at point F slope is positive. Therefore, slope increases at point B with respect to change of x.
Therefore, for minimum value of y, = = > 0.
So, the condition of maxima is = 0 and < 0
And the condition of minimum is = 0 and > 0.
Example: 1. A physical quantity y is given as y = 2x3 – x2 + 2. What are the maximum and minimum value of y?
= 
= 
– 
+ 
= 6x2 – 2x
For maximum or minimum value of y, = 0 or, 6x2 – 2x = 0 or, x(3x – 1) = 0
∴ x = 0 or x = 
.
Now, = 
= 12x – 2
For x = 0, = 0 – 2 = – 2 < 0
Therefore, putting x = 0 on y = 2x3 – x2 + 2, we get the maximum value of y = 2.
For x = , = 12 ×
– 2 = 2 > 0
Therefore, putting x = on y = 2x3 – x2 + 2, we get the minimum value of y = 2
– 
+ 2 = 
– 
+ 2 = 
.
2. Two cars A and B are moving with speed u and v respectively along two mutually perpendicular straight line as shown in figure. What is the time when the distance between two cars is minimum?
Let us consider, two cars A and B initially are at a distance a and b respectively from point O and after time t the distance between two cars is minimum. If L is the minimum distance between two cars then,
L2 = 
+ 
= 
+ 
Or, 2L
= 2(a – ut)(- u) + 2(b – vt)(- v)
As L is the minimum distance, then = 0, therefore, 2(a – ut)(- u) + 2(b – vt)(- v) = 0
Or, 2(a – ut)(- u) = 2(b – vt)v
.
tag:
Share:
koushiadmin
