Maxima and minima in calculus
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 =
and point B represents the minimum value of y when x = 
. Now at point A and B we draw two tangents and the slope of the tangents at those points are 
= 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.
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 = and point B represents the minimum value of y when x = . Now at point A and B we draw two tangents and the slope of the tangents at those points are = 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 D 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 = 2
– 
+ 2. What are the maximum and minimum value of y?
= 
= 
– 
+ 
= 
– 2x
For maximum or minimum = 0 or, – 2x = 0 or, x(3x – 1) = 0
∴ x = 0 or x = 
.
Now = 
= 12x – 2
= 0 – 2 = – 2 < 0 therefore putting x = 0 on y = 2 – + 2, we get the maximum value of y = 2.
= 12 ×
– 2 = 2> 0 therefore putting x = on y = 2 – + 2, we get the maximum 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 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,
= 
+ 
= 
+ 
Or, 2L
= 2 (a-ut)(-u) + 2 (b-vt)
As L is the minimum = 0, then 2 (a-ut) (-u) + 2 (b-vt) (-v) = 0
Or, 2 (a-ut)(-u) = 2(b-vt) v
∴ t = 
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 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,
= +
= +
Or, 2L = 2 (a-ut)(-u) + 2 (b-vt)
As L is the minimum = 0, then 2 (a-ut) (-u) + 2 (b-vt) (-v) = 0
Or, 2 (a-ut)(-u) = 2(b-vt) v
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