Maximum and Minimum Problems - Pre-Calculus
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The profit of a certain cellphone manufacturer can be represented by the function
where 
is the profit in dollars and 
is the production level in thousands of units. How many units should be produced to maximize profit?
The profit of a certain cellphone manufacturer can be represented by the function
where
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We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
What is the critical point of 
. Is it a max or a minimum?
What is the critical point of
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To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
.
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at 
. To get the y value of the point plug x=0 back into the original equation.
thus the point is 
.
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
What is the critical point of 
? Is it a max or minimum?
What is the critical point of
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To find the critical points, we first take the first derivative using the Power Rule.
.
Therefore,
.
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when 
. Now to get the y value of the point we will need to plug in x=0 into the original equation.
. This gives us a critical point at 
.
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
To find the critical points, we first take the first derivative using the Power Rule.
Therefore,
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
Determine whether if there is a maximum or minimum, and location of the point for: 
Determine whether if there is a maximum or minimum, and location of the point for:
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Determine the derivative of this function.
Set the derivative function equal to zero and solve for 
.
Since the parabola has a positive coefficient for 
, the parabola will open upwards, and therefore will have a minimum.
Substitute 
back into the original function and find the y-value of the minimum.
The parabola has a minimum at the point 
.
Determine the derivative of this function.
Set the derivative function equal to zero and solve for
Since the parabola has a positive coefficient for
Substitute
The parabola has a minimum at the point
Find the coordinates of the relative maximum point of the function
.
Find the coordinates of the relative maximum point of the function
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Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Find the 
-coordinates of the possible locations of the relative maxima and minimia of
.
Find the
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We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.
or 
.
We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.
Without solving the problem, determine whether the function 
will have relative maxima or minima and how many of each.
Without solving the problem, determine whether the function
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We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of 
is a positive 6, we also know that the parabola opens upward. Hence, the relative extremum is a minimum.
We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of
The profit of a certain cellphone manufacturer can be represented by the function
where is the profit in dollars and is the production level in thousands of units. How many units should be produced to maximize profit?
The profit of a certain cellphone manufacturer can be represented by the function
where
Tap to see back →
We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
What is the critical point of . Is it a max or a minimum?
What is the critical point of
Tap to see back →
To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
.
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at . To get the y value of the point plug x=0 back into the original equation.
thus the point is .
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
What is the critical point of ? Is it a max or minimum?
What is the critical point of
Tap to see back →
To find the critical points, we first take the first derivative using the Power Rule.
.
Therefore,
.
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when . Now to get the y value of the point we will need to plug in x=0 into the original equation.
. This gives us a critical point at .
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
To find the critical points, we first take the first derivative using the Power Rule.
Therefore,
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
Determine whether if there is a maximum or minimum, and location of the point for:
Determine whether if there is a maximum or minimum, and location of the point for:
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Determine the derivative of this function.
Set the derivative function equal to zero and solve for .
Since the parabola has a positive coefficient for , the parabola will open upwards, and therefore will have a minimum.
Substitute back into the original function and find the y-value of the minimum.
The parabola has a minimum at the point .
Determine the derivative of this function.
Set the derivative function equal to zero and solve for
Since the parabola has a positive coefficient for
Substitute
The parabola has a minimum at the point
Find the coordinates of the relative maximum point of the function
.
Find the coordinates of the relative maximum point of the function
Tap to see back →
Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Find the -coordinates of the possible locations of the relative maxima and minimia of
.
Find the
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We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.
or .
We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.
Without solving the problem, determine whether the function will have relative maxima or minima and how many of each.
Without solving the problem, determine whether the function
Tap to see back →
We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of is a positive 6, we also know that the parabola opens upward. Hence, the relative extremum is a minimum.
We see that the function is a quadratic, so its graph will be a parabola. Thus, we know it will have only one relative max or min. Since the leading coefficient of
The profit of a certain cellphone manufacturer can be represented by the function
where is the profit in dollars and is the production level in thousands of units. How many units should be produced to maximize profit?
The profit of a certain cellphone manufacturer can be represented by the function
where
Tap to see back →
We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
We can determine the production level that maximizes profit by taking the derivative of our function. This can be done using the power rule.
We then set this derivative equal to 0 and solve.
We can factor out our greatest common factor and then divide it out.
Next we can factor.
Therefore, we can solve
We remember that these production levels are in thousands of units.
However, a factory cannot produce negative cellphones, so the maximum profit is obtained when seven thousand units are produced.
What is the critical point of . Is it a max or a minimum?
What is the critical point of
Tap to see back →
To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
.
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at . To get the y value of the point plug x=0 back into the original equation.
thus the point is .
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
To find the critical points, we first take the first derivative using the Power Rule:
Therefore,
To find the critical point we need to set the derivative equal to zero and solve for x.
Doing so this gives the critical point at
Is it a max or min?
Take the second derivative
It is a minimum since the second derivative is positive.
What is the critical point of ? Is it a max or minimum?
What is the critical point of
Tap to see back →
To find the critical points, we first take the first derivative using the Power Rule.
.
Therefore,
.
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when . Now to get the y value of the point we will need to plug in x=0 into the original equation.
. This gives us a critical point at .
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
To find the critical points, we first take the first derivative using the Power Rule.
Therefore,
From here we need to set the derivative equal to zero and solve to x.
The critical point occurs when
To find if it is a max or min, we take the second derivative
Since the second derivative is negative, it is a maximum.
Determine whether if there is a maximum or minimum, and location of the point for:
Determine whether if there is a maximum or minimum, and location of the point for:
Tap to see back →
Determine the derivative of this function.
Set the derivative function equal to zero and solve for .
Since the parabola has a positive coefficient for , the parabola will open upwards, and therefore will have a minimum.
Substitute back into the original function and find the y-value of the minimum.
The parabola has a minimum at the point .
Determine the derivative of this function.
Set the derivative function equal to zero and solve for
Since the parabola has a positive coefficient for
Substitute
The parabola has a minimum at the point
Find the coordinates of the relative maximum point of the function
.
Find the coordinates of the relative maximum point of the function
Tap to see back →
Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Let's find the first derivative to locate the relative maxima and minima.
Now we set it equal to zero to find the x values of these critical points.
So the equation is 0 where x is -2, 0, or 5. Now let's find the second derivative so that we know which of these locations are maxima and which are minima.
So the function has a relative maximum at x=-5.
So the function has a relative minimum at x=0.
So the function has a relative maximum at x=2.
Thus there is only one relative minimum in this function, and it occurs at x=0. We need to plug this into the original function to find the y-coordinate of the point.
So our point is (0,8).
Find the -coordinates of the possible locations of the relative maxima and minimia of
.
Find the
Tap to see back →
We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.
or .
We need to find out where the first derivative is equal to zero to find the locations of the possible maxima and minima.