Finding extreme points under constraint

## Finding extreme points under constraint with an example There are many cases, especially in functions related to the business sector, where the extreme point is at values ​​of x and y that are not within the firm’s reach, or in other words, some of the values ​​of the variables are irrelevant. ### Example 1 An entrepreneur established a kettle factory. Production is carried out using workers and machines. The entrepreneur discovered that he could predict the monthly production volume using a production function whose symbol and components are: – **Monthly production function**:
– `x` denotes the number of verbs
– `y` denotes the number of machines
– `P` (short for Production) – represents the number of kettles. Market prices are as follows: – Wage: 4,000 NIS per month
– Machine operation: 1,000 NIS per month
– Price of a kettle: 100 NIS The factory economist was asked to find at what production mix (of workers and machines) the maximum profit should be obtained, under the constraint that the expenditure budget will be 40,000 NIS per month. For the sake of simplicity, we assume that the factory’s expenses include only labor and machine operation. ### A basket of production factors for any combination of workers and machines is called a basket of production factors or basket for short. Within the framework of the constraint (in the example), the cost of the basket is 40,000 NIS. ### Constraint – Extension The constraint is expressed in the fact that not all possible production baskets are relevant, but only the baskets that cost 40,000 NIS. ### Presenting the constraint as a function on the axis plane The requirement that the cost of the basket be 40,000 NIS can be presented using equality 1. – **Equality 1**: `40,000=4,000x+1,000y` Only baskets that satisfy equality 1 are relevant. If we divide equality 1 by 1,000 and isolate y, we obtain equality 2: – **Equality 2**: `y=40-4x` Equation 2 represents a function of a straight line. ### Properties of the function 1. Intersects the y-axis at 40.
2. Slope: 4. The cost of all baskets along the function path is exactly 40,000 NIS. For example, basket A is worth 40,000 NIS (where `40,000 =40*1,000`), just like basket B (`10*4000`) and basket C (`16*1000+6*4000`). ### Possibility of finding the maximum profit also using the production function P(x,y) Since we assumed that the factory’s expenses will be 40,000 NIS, then the factory will maximize its profits where the production function will reach a maximum. In other words, the basket of production factors that yields the maximum output and is within the budget also yields the maximum profit. ### Finding the maximum profit – Introduction If we place the coordinate points along the budget line in the production function, we obtain a strip (which is part of the production function). The highest point on the strip is the maximum extreme point. ### Placing the coordinate points of the budget line in the production function At all coordinate points that are on the budget line, the value of y can be displayed using `[40-4x] `, since `y=40-4x`. That is, for every y there is an alternative in the values ​​of x that is `[40-4x] `. When we substitute the alternative ` in the production function instead of y[40-4x] `, we get: – `P(x,y) = 20xy = 20x[40-4x] = 800x-80x^2` The function we obtained represents the strip above the budget constraint line. ### Finding a maximum absolute extremum point in a strip We have already learned that an extremum point in any strip can only be found in two places: 1. The edges of the strip – at one or both ends.
2. Where the derivative is equal to 0 and provided that it is an extremum point and not an inflection point.

#### Testing technique We will check at what value of x the band above the budget line reaches a maximum. 1. **Checking the result at the ends of the band**
– The form of the production function is `P(x)=800x-80x^2`.
– When we set `x = 0`, the result is: 0 kettles = `P(x)`.
– When we put `x = 10` in it, the result is: 0 kettles = `P(x)`. 2. **Checking the result along the strip**
– We will check using the first derivative whether there is a point on the strip where the slope is 0.
– The derivative is: `P'(x)=800-160x`.
– The derivative is equal to 0 when: `x=5`.
– The result of the function when `x=5` is 2000 kettles: `P(5)=800*5-80*5^2=2000`.
– We will check using the second derivative whether this is an extreme point or an inflection point.
– The second derivative is: `P”(x)=-160`.

### Test results 1. **Left end** (when `x=0`) The result of the function is: 0 kettles.
2. **Right edge** (when `x=10`) The result of the function is: 0 kettles.
3. **Calculating the result of the function at the x value where the derivative is equal to 0**:
– The function is: `P(x)=800x-80x^2`
– The derivative is: `P'(x)=800-160x`
– The derivative equals 0 when `x=5`.
– The result of the function when `x=5` is 2000 kettles `(800*5-80*5^2=2000)`.
– At about `x=5` the function result is higher than at the ends.
4. **We will check that this is an extreme point and not an inflection point** – The result of the second derivative is `-160[P”(x)=-160] `. ### Conclusion The production function reaches a maximum (2000 kettles) when the factory employs 5 workers. ### Finding the number of machines at the maximum point We calculate the number of machines from the budget line equation (when we put 5 in place of x). – Budget line equation: `y=40-4x`
– When you put `5 = x` in it, you get: 20 machines `(20=40-20)` ### Conclusion The factory maximizes profits when it employs 5 workers and 20 machines. In this scenario, the activity data are as follows: – Quantity of kettles produced per month: 2000 units
– Revenue: 200,000 NIS
– Profit: 160,000 NIS — ### Example 2 – The Kettle Factory Year After Year After Year, the production function of the kettle factory improved and became: `P(x,y)=40x+100y` (`P` – the number of kettles produced). The rest of the market data (wages, machine operating costs, and kettle price) did not change. The constraint of staying within the budget of 40,000 NIS is still in effect. Therefore, the budget equation and budget line have not changed. We calculate the results of the production function above the budget line using only the values ​​of x. To do this, we substitute its alternative `[40-4x] ` and we get: – `P(x)=40x+100*((y),(40-4x))`
– `=40x+4000-400x`
– `=4000-360x` The function we got is the band function above the budget line. ### Its characteristics – Straight line. – Negative slope. ### Finding an extreme point in a band #### The results of the function at the ends of the band – **Left end** (when `x=0`) The result of the function is 4000 kettles. – **Right end** (when `x=10`) The result of the function is 400 kettles.

#### Calculating the result of the function at the x value where the derivative equals 0 – the form of the function: `f(x)=4000-360x`
– The form of the derivative: `f'(x)=-360` The meaning of the derivative is that the slope at each point on the strip is -360 (=straight line). That is, there is no extreme point along the strip. ### Conclusion The factory maximizes its profits when `x=0`, that is, when it employs no workers, but only 40 machines. At the extreme point in this scenario, the activity data are as follows: – Quantity of kettles produced: 4,000 units (compared to 2,000 in the previous year)
– Revenue: 400,000 NIS (compared to 200,000 in the previous year)
– Profit: 360,000 NIS (compared to 160,000 in the previous year) If the factory had employed only workers (10 workers and 0 machines), then the activity data would have been as follows: – Quantity of kettles produced: 400 units `(P(10)=4000-360*10=)`
– Revenue: 40,000 NIS `(= 400*100 NIS)`
– Profit: 0 NIS (expense budget = 40,000 NIS) — ### Example 3 We are looking for an extreme point for the function `f(x,y)=x*y` under the constraint that the function is relevant only when the x values ​​are the same as the y values, or if you prefer, only at coordinates where `x=y`. In the following diagram, the diagonal line (1), whose slope is 1, shows all pairs in the x,y plane that satisfy the constraint. The relevant function is the strip above line (1). If we substitute in the function, instead of y, its alternative in values ​​of x, which is `[x]`, the function will become a function of x only, whose symbol and form are `f(x)=x^2`. #### Finding an extreme point 1. The function has no starting and ending points, it descends from the sky when `x =-∞`, touches the plane of the axes when x= 0 and rises back to the sky when `x = ∞`.
2. Find if and where its derivative equals 0.
– The function is `f(x,y) =xy`, and after setting x=y it becomes `f(x)=x^2`.
– The derivative is: `f'(x)=2x`.
– The derivative equals 0 when `x=0`.
3. We will check using the second derivative if this is an extreme point and which one? (maximum or minimum). – The result: `f”(x)=2` (a convex curve, whose slopes increase by 2). – And the conclusion: this is a minimal extreme point. We find the partner of `x=0` on line (1). The result: `0= y (y=x)`. That is, the strip reaches a minimum above the coordinate point (0,0) and the height of the function at the minimal extreme point is 0. — ### Example 4 We are looking for an extreme point for the (spatial) function `f(x,y)=xy` from the previous example, but under a different constraint. The constraint is that the spatial function is relevant only for values ​​of y that satisfy the equation `y=4-x`. In other words, the function is relevant only when the partner of each x is `[4-x] `. In the following diagram, the diagonal line (1) shows all pairs that satisfy the constraint.

#### Finding an extremum 1. The function has no starting and ending point.
2. Find if and where its derivative equals 0.
– The derivative is: `f'(x) = 4-2x`
– The derivative equals 0 when `x=2`.
3. We will check using the second derivative whether this is an extreme point and which one?
– The result: `f”(x) = -2` (a concave curve, whose slopes decrease by 2).
– And the conclusion: a maximum extreme point. At about `x=2` the function is at a maximum extreme point. We find the partner of x=2 on line (1) in the space of the axes. The result is `y=4-2=2`. That is, the strip reaches a maximum above the coordinate point (2,2) (point A in the previous diagram, viewed from above).