If f(x) is a continuous function on a closed interval, f has both an absolute maximum value and an absolute minimum value. Before we get into this, let’s go over some other basic function analysis:
A critical point is the point on the function where the derivative is 0, commonly the point where the function changes direction, but not always.
Relative maximum and minimum values occur at critical points. This means that that point is the highest or lowest point in the area- meaning that points on the left or right of it do not have a greater/lower value.
You can use the first derivative test to determine if a point is a relative max or relative min. This means you find the first derivative function, and plug in points around your value to see if the sign changes.
You can also use the second derivative test. The second derivative will tell you the concavity of your original function. So, after you take the derivative of your first derivative expression, plug in your value to see if you get a negative or positive value. If the second derivative test gives you a negative value, then this is a maximum value, because it will be concave down. If you get a positive value, this is a minimum value. If you get 0, the test is inconclusive and you should use the first derivative test.
Inflection points are any point on the function where the concavity of the function changes.
Identify Q1 and Q2, your two quantities. If dQ1/dt is the given, then you are solving for dQ2/dt.
Example:
You’re building a pen onto a wall, meaning you are only adding three sides. You have 60m of fencing, find dimensions of the pen that will maximize the area.
A critical point is the point on the function where the derivative is 0, commonly the point where the function changes direction, but not always.
Relative maximum and minimum values occur at critical points. This means that that point is the highest or lowest point in the area- meaning that points on the left or right of it do not have a greater/lower value.
You can use the first derivative test to determine if a point is a relative max or relative min. This means you find the first derivative function, and plug in points around your value to see if the sign changes.
You can also use the second derivative test. The second derivative will tell you the concavity of your original function. So, after you take the derivative of your first derivative expression, plug in your value to see if you get a negative or positive value. If the second derivative test gives you a negative value, then this is a maximum value, because it will be concave down. If you get a positive value, this is a minimum value. If you get 0, the test is inconclusive and you should use the first derivative test.
Inflection points are any point on the function where the concavity of the function changes.
Identify Q1 and Q2, your two quantities. If dQ1/dt is the given, then you are solving for dQ2/dt.
- Draw it out.
- (permanent relationship) Create an equation that models the mathematical relationship between the two quantities. Make sure this
equation remains true any changes made to Q1 or Q2.
- Differentiate the equation with respect to time so that it now contains dQ1/dt and dQ2/dt.
- Substitute the instantaneous values that were given in the original problem, including dQ1/dt.
- Solve for any other unknowns before solving for the roc of q2.
- SolvefordQ2/dt
Example:
You’re building a pen onto a wall, meaning you are only adding three sides. You have 60m of fencing, find dimensions of the pen that will maximize the area.
Optimization is also used in companies to maximize profit. Remember: Profit = Revenue - Cost.
Say a firm determines that x units of its product can be sold daily at p dollars per unit, using this equation:
x = 1000-p
The cost of making x units per day is shown by : C(x)=3000+20x
Find the number of units that should be produced to maximize profit, what that profit is, and what the price per unit is.
Say a firm determines that x units of its product can be sold daily at p dollars per unit, using this equation:
x = 1000-p
The cost of making x units per day is shown by : C(x)=3000+20x
Find the number of units that should be produced to maximize profit, what that profit is, and what the price per unit is.