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Optimization in Calculus

April 10th, 2009 |

Author: Merrin

You can gather some information from the derivatives of a function. If the derivative is less than zero at, to the left, and to the right of a point then the function is decreasing there. Similarly you can tell when a function is increasing if the sign is positive. What happens if the derivative of a function is zero at a point? Well if to the right it is positive and to the left it is positive, then you know that point is a local minimum. Similarly you can find a local maximum with the signs reversed. If you don’t like the left and right business you can always take the second derivative. If the second derivative is positive at a point where the first derivative is zero, then that point is a local minimum. This is because it means the derivative is increasing to the left and right of that point. Similarly you can find a local maximum with the signs reversed.

If you want to find the absolute maximum or absolute minimum of a function over a region this is called an optimization problem. In order to solve this problem you have to identify all the so called critical points and evaluate the function at these points and compare the value. The endpoints of the domain of the function are critical points. Points where the derivative does not exist or equals zero are also critical points. Many optimization word problems can be formulated into a function to be optimized. Set the derivative of this function equal to zero and find the critical points and generally this is enough to solve the problem with the above considerations also taken into account.

Example 1. A farmer wants to build a small rectangular pigpen alongside the side of his barn. His barn is 100m long and he has 80m of fence and two posts. How far away from the side of his barn should he place the posts to get the maximum area rectangular pigpen?

Solution 1. We can make a diagram of the pigpen as follows. Since the perimeter of the pigpen not touching the barn must equal the length of the fence we have

$2x+y=80$

The area of the pigpen is given by

$A=xy=x(80-2x)$

We can maximize the area by calculating

${\cfrac{dA}{dx} } = {0=-4x+80}$

${x} = {20}$

The other critical points are x = 0 and x = 40 but those dimensions give zero area. The pigpen should be built to be 20m by 40m.

Example 2. You have a piece of cardboard that is 20 cm by 30 cm. What is the largest volume box you can make by cutting out four squares in the corners and then folding the sides upwards?

Solution 2. The volume of the box in the above diagram is

${V(x)}={x(20-2x)(30-2x)}$

$= {x(600-60x-40x-4x^{2} )}$

$={+4x^{3} -100x^{2} +600x}$

We can maximize the volume by finding $latex V’(x)=0$.

${\cfrac{dV}{dx} =+12x^{2} -200x+600=0}$

Solving the quadratic equations we have

${x=\cfrac{25}{3} \pm \cfrac{5}{3} \sqrt{7} } \\\\\\ {x_{+} =12.74...\quad x_{-} =3.92....} \\\\\\ {V(x_{-} )=1056\, cm^{3} }$

The critical points are $latex x=0,10,x_{+} ,\, {\rm and}\, x_{-} $. $latex V(0)=V(40)=0\, $and $latex x_{+} $ is too large to be possible.

Example 3. A lifeguard has his station right at the water. He sees a swimmer in trouble a distance 50m away, 30m from the shore, and 40m as measured along the sand and perpendicular to the swimmer. The lifeguard can run on the sand at a speed of 8m/s and he can swim in the water at a rate of 2m/s. What is the optimal path the lifeguard should take so as to reach the troubled swimmer in the least amount of time?

Solution 3. The first thing we should do is set up a diagram.

The optimal path is equivalent to finding x the point at which he should enter the water and then swim in a straight line to the troubled swimmer. The next step is to write down an equation for the time of each path and minimize it.

$t=\cfrac{x}{8} +\cfrac{y}{2}$

Now we know that

$y^{2} =30^{2} +(40-x)^{2}$

So our new function to be minimized is

$t=\cfrac{x}{8} +\cfrac{\sqrt{30^{2} +(40-x)^{2} } }{2}$

Our endpoints are x = 0 and x = 40. We know from physical reasons these are not the minimal times.

${\cfrac{dt}{dx} =0=\cfrac{1}{8} +\cfrac{1}{2} \cfrac{(x-40)}{\sqrt{(x-40)^{2} +30^{2} } } } \\ \\ {x=(40-2\sqrt{15} )=32.25}$

Plugging in this coordinate we can calculate the least time to reach the swimmer.

$t=19.52s \quad \text{when} \quad x=32.25m$

Example 4. A light ray moves through medium 1 with a speed $latex \cfrac{c}{n_{1} } $ and moves through a medium 2 with speed $latex \cfrac{c}{n_{2} } $. If medium one is a horizontal slab connected to medium 2 we can describe how a light ray moves from a point A to point B according to what is known as Fermat’s principle of least time. Calculate the time it takes for light to travel straight line paths that bend at the interface between the two slabs.

Solution 4. The first thing we do is to draw a diagram so we can set up the problem. The second step is to try to set up an equation for what we need to optimize. Let the light start at the origin in medium 1 and the target location is (x,y) in medium 2. Let the light cross the interface with coordinate \textit{x1}.

${t=\cfrac{n_{1} }{c} \sqrt{x_{1}^{2} +(y/2)^{2} } +\cfrac{n_{2} }{c} \sqrt{(x-x_{1} )^{2} +(y/2)^{2} } } \\ {\cfrac{dt}{dx_{1} } =0=\cfrac{n_{1} }{c} \cfrac{x_{1} }{\sqrt{x_{1}^{2} +(y/2)^{2} } } +\cfrac{n_{2} }{c} \cfrac{(x_{1} -x)}{\sqrt{(x-x_{1} )^{2} +(y/2)^{2} } } } \\ {n_{1} \cfrac{x_{1} }{\sqrt{x_{1}^{2} +(y/2)^{2} } } =n_{2} \cfrac{(x-x_{1} )}{\sqrt{(x-x_{1} )^{2} +(y/2)^{2} } } } \\ {n_{1} \sin \theta _{1} =n_{2} \sin \theta _{2} }$

Fermat’s principle of least time gives us Snell’s law, which relates the angle of incidence and angle of refraction according to the index of refraction of the two mediums.

Example 5. A triangle inscribed in a circle has its base along a diameter of the circle. Find the maximum area of the inscribed triangle

Solution 5. The first thing to do is to draw a graph and choose variables.

The area of the triangle is given as $latex A=bh=\cfrac{1}{2} (2R)R\sin \theta $. Next we differentiate to find the maximum area.

${\cfrac{dA}{d\theta } =\cfrac{1}{2} DR\cos \theta =0}$