Optimization is a process of finding an extreme value (either maximum or minimum) under certain conditions.

**Applications in
Optimization**

Optimization
is a process of finding an extreme value (either maximum or minimum) under
certain conditions.

A
procedure for solving for an extremum or optimization problems.

**Step 1 : **Draw an appropriate figure and label
the quantities relevant to the problem.

**Step 2 : **Find a experssion for the quantity
to be maximized or minimized.

**Step 3 : **Using the given conditions of the
problem, the quantity to be extremized .

**Step 4 : **Determine the interval of possible
values for this variable from the conditions given in** **the problem.

**Step 5 : **Using the techniques of extremum
(absolute extrimum, first derivative test or second** **derivative test) obtain the maximum or minimum.

** **

**Example 7.62**

We have
a 12 unit square piece of thin material and want to make an open box by cutting
small squares from the corners of our material and folding the sides up. The
question is, which cut produces the box of maximum volume?

**Solution**

Let *x* = length of the cut on each side of
the little squares.

* V* = the
volume of the folded box.

The
length of the base after two cuts along each edge of size *x* is 12 âˆ’ 2*x* . The depth of the box after folding is *x* , so the volume is *V* =
*x* Ã— (12 âˆ’ 2*x*)^{2} . Note that, when *x* = 0 or 6 , the volume is zero and hence there cannot be a
box. Therefore the problem is to maximize, *V*
=
*x* Ã— (12 âˆ’ 2*x* )^{2} , *x* âˆˆ(0, 6).

Now, dV /
*dx *= (12 âˆ’ 2x)^{2} âˆ’ 4x(12 âˆ’
2*x*)

= (12 âˆ’ 2x )(12 âˆ’ 6x).

dV / *dx* = 0 gives the stationary numbers* x *= 2, 6 . Since 6âˆ‰(0,6) the only stationary number is
at* x *= 2 âˆˆ(0, 6) . Further, dV* / dx*
changes its sign from postive to negative
when passing through* x *= 2 .
Therefore at* x *= 2 the volume *V* is local maximum. The local maximum
volume value is *V* =
128 units. Hence the maximum cut can only be 2 units.

** **

Find the
points on the unit circle *x*^{2}
+
*y*^{2} =
1 nearest and farthest from (1,1) .

**Solution**

The
distance from the point (1,1) to any point (*
x *, y) is *d* = âˆš{(x âˆ’1)^{2}
+ (* y *âˆ’1)^{2}} . Instead of extremising
*d*, for convenience we extremise D = *d*^{2} = (*x* âˆ’ 1)^{2} + (*y *âˆ’1)^{2},
subject to the condition *x*^{2}
+* y*^{2} =1. Now, dD/*dx* = 2(*x* âˆ’ 1) + 2(* y *âˆ’1) Ã— *dy*/*dx*
, where the *dy*/*dx* will be computed by differentiating *x*^{2} +* y*^{2}
= 1 with respect to* x *. Therefore, we
get 2x + 2*y dy*/*dx* = 0 which gives us *dy*/*dx* = âˆ’* x/y *.

Since ( *x* , *y*)
lie on the circle *x*^{2}* *+* y*^{2}* *=1, we get 2*x*^{2} =1 which gives *x* = Â± 1/âˆš2 . Hence the points at which the extremum distance occur
are, (1/âˆš2, 1/âˆš2), (-1/âˆš2, -1/âˆš2).

To find
the extrema, we apply second derivative test. So,

This
implies the nearest and farthest points are (1/âˆš2, 1/âˆš2) and (-1/âˆš2, -1/âˆš2) respectively.

Therefore,
the nearest and the farthest distances are respectively âˆš2 - 1 and âˆš2 + 1.

** **

A steel
plant is capable of producing *x*
tonnes per day of a low-grade steel and *y*
tonnes per day of a high-grade steel, where *y*
=
40âˆ’5*x * / 10-*x or *[].
If the fixed market price of low-grade steel is half that of high-grade steel,
then what should be optimal productions in low-grade steel and high-grade steel
in order to have maximum receipts.

Let the
price of low-grade steel be â‚¹*p* per tonne. Then the price of
high-grade steel is â‚¹2*p*
per tonne.

The
total receipt per day is given by R = p*x*
+ 2*py* = *px* + 2p[40âˆ’5*x
* / 10-*x*]. Hence the problem is to maximise *R* . Now, simplifying and differentiating *R* with respect to *x* , we
get

and
hence R will be maximum. If* x *= 10-2âˆš5
then* y *= 5 âˆ’ âˆš5 . Therefore the steel
plant must produce low-grade and high-grade steels respectively in tonnes per
day are

10-2âˆš5
and 5 âˆ’ âˆš5.

Prove
that among all the rectangles of the given area square has the least perimeter.

**Solution**

Let *x* , *y*
be the sides of the rectangle. Hence the area of the rectangle is *xy* = *k*
(given). The perimeter of the rectangle *P*
is 2(*x* + *y*)
. So the problem is to minimize 2(*x* +
*y*) suject to the condition *xy* = *k*
. Let P (*x*) = 2* *(*x *+ *k/x*).

* P*
â€²(*x*) = 0 â‡’* x *= Â±âˆšk.

As x,* y *are sides of the rectangle,* x *= âˆšk is a critical number.

Now, *P* â€²â€² (x) = 4*k*/* x*^{3} and P â€²â€²
( âˆšk ) > 0 â‡’ P ( x)
has its minimum value at* x *= âˆšk .

Substituting* x *= âˆš*k* in *xy* = *k* we get* y *= âˆš*k* . Therefore the minimum
perimeter rectangle of a given area is a square.

Tags : Applications of Differential Calculus | Mathematics , 12th Maths : UNIT 7 : Applications of Differential Calculus

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