Minimization and maximization refresher
The fundamental idea which makes calculus useful in understanding problems of maximizing and minimizing things is that at a peak of the graph of a function, or at the bottom of a trough, the tangent is horizontal. That is, the derivative $f'(x_o)$ is $0$ at points $x_o$ at which $f(x_o)$ is a maximum or a minimum.
Well, a little sharpening of this is necessary: sometimes for either natural or artificial reasons the variable $x$ is restricted to some interval $[a,b]$. In that case, we can say that the maximum and minimum values of $f$ on the interval $[a,b]$ occur among the list of critical points and endpoints of the interval.
And, if there are points where $f$ is not differentiable, or is discontinuous, then these have to be added in, too. But let's stick with the basic idea, and just ignore some of these complications.
Let's describe a systematic procedure to find the minimum and maximum values of a function $f$ on an interval $[a,b]$.
- Solve $f'(x)=0$ to find the list of critical points of $f$.
- Exclude any critical points not inside the interval $[a,b]$.
- Add to the list the endpoints $a,b$ of the interval (and any points of discontinuity or non-differentiability!)
- At each point on the list, evaluate the function $f$: the biggest number that occurs is the maximum, and the littlest number that occurs is the minimum.
Find the minima and maxima of the function $f(x)=x^4-8x^2+5$ on the interval $[-1,3]$. First, take the derivative and set it equal to zero to solve for critical points: this is $$4x^3-16x=0$$ or, more simply, dividing by $4$, it is $x^3-4x=0$. Luckily, we can see how to factor this: it is $$x(x-2)(x+2)$$ So the critical points are $-2,0,+2$. Since the interval does not include $-2$, we drop it from our list. And we add to the list the endpoints $-1,3$. So the list of numbers to consider as potential spots for minima and maxima are $-1,0,2,3$. Plugging these numbers into the function, we get (in that order) $-2, 5, -11, 14$. Therefore, the maximum is $14$, which occurs at $x=3$, and the minimum is $-11$, which occurs at $x=2$.
Notice that in the previous example the maximum did not occur at a critical point, but by coincidence did occur at an endpoint.
You have $200$ feet of fencing with which you wish to enclose the largest possible rectangular garden. What is the largest garden you can have?
Let $x$ be the length of the garden, and $y$ the width. Then the area is simply $xy$. Since the perimeter is $200$, we know that $2x+2y=200$, which we can solve to express $y$ as a function of $x$: we find that $y=100-x$. Now we can rewrite the area as a function of $x$ alone, which sets us up to execute our procedure: $$area = xy=x(100-x)$$ The derivative of this function with respect to $x$ is $100-2x$. Setting this equal to $0$ gives the equation $$100-2x=0$$ to solve for critical points: we find just one, namely $x=50$.
Now what about endpoints? What is the interval? In this example we must look at ‘physical’ considerations to figure out what interval $x$ is restricted to. Certainly a width must be a positive number, so $x>0$ and $y>0$. Since $y=100-x$, the inequality on $y$ gives another inequality on $x$, namely that $x <100$. So $x$ is in $[0,100]$.
When we plug the values $0,50,100$ into the function $x(100-x)$, we get $0,2500,0$, in that order. Thus, the corresponding value of $y$ is $100-50=50$, and the maximal possible area is $50\cdot 50=2500$.
- Olivia has $200$ feet of fencing with which she wishes to enclose the largest possible rectangular garden. What is the largest garden she can have?
- Find the minima and maxima of the function $f(x)=3x^4-4x^3+5$ on the interval $[-2,3]$.
- The cost per hour of fuel to run a locomotive is $v^2/25$ dollars, where $v$ is speed, and other costs are $100 per hour regardless of speed. What is the speed that minimizes cost per mile?
- The product of two numbers $x,y$ is 16. We know $x\geq 1$ and $y\geq 1$. What is the greatest possible sum of the two numbers?
- Find both the minimum and the maximum of the function $f(x)=x^3+3x+1$ on the interval $[-2,2]$.