Page 345 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes 25 26
When x is slightly greater than say , we have
16 16
1 16 2
2 26 5 = 0.392 0.4 0

dy 25
Since the sign of changes from positive to negative as we pass through the point x , the function
dx 16
has a maxima at this point

Example: A rectangular area is to be marked off as a chicken run with one side along an
existing wall. The other sides are marked by wire netting of which a given length is available.
Show that the area of the run is maximum if one side is twice the other.

Solution:
Let x be the length and y be the breadth of rectangle. Also let l be the length of wire.
We can write l = x + 2y, or x l 2y .

The area of the rectangle,
A = x.y = l 2y y ly 2y 2
We want to find y so that A is maximum.
dA l
dy = l 4y 0 or y 4 , for maxima.

Second order condition:
2
d A = therefore A is maximum when y l
dy 2 4 .

l l
Also x = l 2y l .
2 2
Thus A is maximum when one side is taken as twice the other.

Example: An open box is constructed by removing a small square of side x cms from
each corner of the metal sheet and turning up the edges. If the sheet is a square with each side
equal to L cms, find the value of x so that volume of the box is maximum. Also find the largest
volume of the box.

Figure 13.9
Wall

y y
x
Fig. 5.9
Solution:

After a square of side x cms is removed from each corner, the base of the box will be a square
with each side = L 2x .

Volume of the box V = x L 2x 2

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