# Extreme value problem

One type of task that uses differential calculus are optimizations or **extreme value problems**. An extreme value problem is a kind of optimization problem but only with one constraint. It can be both a maximization problem and a minimization problem.

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### Hint

Extreme value problems exist when a target size (e.g., surface area, volume, profit, ...) is to become maximum or minimum. This condition is then the

**main condition**.!

### Remember

In the case of extreme value problems, a function (

**the objective function**) is set up from a**main condition**and a**constraint**, whose extreme values are searched for.i

### Method

- Main condition
- Constraint
- Set up objecitve function
- Calculate extreme values of the objective function
- Calculate missing sizes

### Example

The largest possible rectangle area has to be limited with an 800 m fence. Calculate the size of both sides and the area.

#### Main condition

*The area of the rectangle should be maximized. Therefore, this is the main condition and depends on two variables $a$ and $b$.*

$A(a, b)=a\cdot b$#### Constraint

*There are only 800 m of fence available, which limits the area. This is the perimeter of the rectangle.*

$P=2a+2b$

$800=2a+2b$#### Set up objective function

*To link both conditions, the constraint is changed to a variable.*

$800=2a+2b\quad|-2b$

$800-2b=2a\quad|:2$

$a=\frac{800-2b}2$ $=400-b$*Now this must be used in the main condition and you get the objective function, which is dependent on only one variable.*

$A(a, b)=a\cdot b$

$A(b)=(400-b)\cdot b$ $=400b-b^2$-
#### Calculate extreme values of the objective function

*Now you can (as with other functions too) calculate the extreme values of the objective function.*

$A(b)=400b-b^2$

$A'(b)=400-2b$

$400-2b=0\quad|-400$

$-2b=-400\quad|:(-2)$

$b=200$*With the second derivation, it is still checked whether the result is actually a maximum point, since the area should be***maximum**.

$A''(b)=-2$

$A''(200)=-2<0$ => Maximum point -
#### Calculate missing sizes

$b=200m$*From the (converted) constraint you can now calculate $a$.*

$a=400-b$

$a=400-200=200m$*From the main condition (alternatively also with the objective function) the area $A$ can be calculated.*

$A(a, b)=a\cdot b$

$A(a, b)=200m\cdot 200m=40.000m^2$