12.3: Optimization

Solution

Here we are again trying to find the maximum possible area, so our optimizing function is \(A(w,d)=w\cdot d\) where \(w\) is the width and \(d\) is the depth in the diagram above.

The constraint is that the amount of fence material must total to 60$ meters. Since we need 3 walls of length \(d\) and one of length \(w\), we need \(3d+w=60\)

Notice that it is easy to solve for either variable here in terms of the other, and it does not matter which you pick as it will ultimately lead to the same final answer. To make our lives easier we will just solve for \(w\), giving \(w=60-3d\)

Now substituting this into \(A\) we get \(A(d)=(60-3d)\cdot d=60d-3d^2\)

This function has a natural domain since \(w\ge 0\) and \(d\ge 0\) must hold, so we get \(d\in [0,20]\)

Now we can differentiate, giving \(A'(d)=60-6d\)

This always exists, so our critical points are when \(A'(d)=0\), giving only \(d=10\)

We check our critical points and endpoints to find the maximum for \(A\):

\(A(0)=60\cdot 0-3\cdot 0^2=0 \)
\(A(10)=60\cdot 10-3\cdot 10^2=200\)
\(A(20)=60\cdot 20-3\cdot 20^2=0 \)

Clearly the maximum possible area is 200 meters\(^2\), but the question asks for the width.

We can plug in \(d=10\) to the equation \(w=60-3d\) to get \(w=30\) meters.

Therefore our answer is the maximum area occurs when the width is 30 meters.

Solution

Remember a cylinder with base radius \(r\) and height \(h\) has a volume of \(V=\pi r^2 h\) and a surface area of \(S=2\pi r^2+2 \pi rh\)

We want to find the smallest surface area, so that should be our optimizing function: \(S(r,h)=2\pi r^2+2\pi rh\)

Our constraint is that we want the volume to be \(250\pi\), therefore \(250\pi=\pi r^2h\)

We can solve for \(h\) more easily than \(r\) here, giving \(h=\displaystyle{\frac{250}{r^2}}\)

We substitute this into our optimizing function to get:

\(\displaystyle{S(r)=2\pi r^2+2\pi r\cdot \frac{250}{r^2}=2\pi r^2+500\pi r^{-1}}\)

The domain here is again \(r,h\ge 0\) must hold, but this only tells us \(r\in[0,\infty)\).

Differentiating this gives \(S'(r)=4\pi r-500\pi r^{-2}=\displaystyle{\frac{4\pi r^3-500\pi}{r^2}}\)

Thus \(S'(r)=0\) when \(4\pi r^3-500\pi=0 \longrightarrow r=5$\)

and \(S'(r)=\)DNE when \(r^2=0 \longrightarrow r=0\)

Since our domain is not a closed interval, we make a sign chart:

Our local minimum must therefore occur at \(r=5\) yards. This question asked us to find the smallest possible surface area, so we plug this into \(S(r)\) to get:

\(S(5)=2\pi\cdot 5^2+500\pi \cdot 5^{-1}=50\pi+100\pi=150\pi\)

Our answer is officially that the minimum surface area is \(150\pi\) yards\(^2\).

Solution

Here it helps to draw a sketch of what is going on:

Our goal is to minimize the length of the purple line connecting \(\textcolor{red}{(x,y)}\) to \(\textcolor{brown}{(2,1)}\).

We can find this length by using the distance formula: \(\textcolor{purple}{d(x,y)=\sqrt{(x-2)^2+(y-1)^2}} \)

The constraint is that the point \(\textcolor{red}{(x,y)}\) must be on the line \(y-2x=1\), which we can easily solve for \(y\) in to get: \(y=2x+1\)

Now we substitute that into our optimizing function \(d\) to get:

\(d(x)=\sqrt{(x-2)^2+((2x+1)-1)^2}=\sqrt{5x^2-4x+4}\)

The domain is clearly \((-\infty,\infty)\), but you can check by hand if needed.

Next, we find \(d'(x)=\displaystyle{\frac{1}{2}(5x^2-4x+4)^{-\frac{1}{2}}\cdot (10x-4)=\frac{5x-2}{\sqrt{5x^2-4x+4}}}\)

We have \(d'(x)=0\) when \(5x-2=0 \longrightarrow \displaystyle{x=\frac{2}{5}}\)

And \(d'(x)=\)DNE when \(\sqrt{5x^2-4x+4}=0\) (which has no real solutions, so never)

A quick check of the sign chart confirms we indeed have a minimum at \(x=\displaystyle{\frac{2}{5}}\):

We want to find the closest point, so we need to find \(y\) next:

Since \(y=2x+1\), we plug in \(\displaystyle{x=\frac{2}{5}}\) to get \(\displaystyle{ y=\frac{9}{5}}\)

So our final answer is the closest point is at \(\displaystyle{\left(\frac{2}{5},\frac{9}{5}\right)}\)

Solution

Once again we start with a sketch to help see what is going on:



We can parameterize the rectangle in terms of the corner that lands in quadrant one at \((x,y)\). Moving this point along the parabola will create infinitely many different rectangles. We want to maximize the area of the blue rectangle, which has a base of \(2x\) and a height of \(y\). This gives our optimizing function as \(A(x,y)=2xy\)

The constraint is that the point \((x,y)\) must be on the parabola \(y=12-x^2\), (which is already set up to be substituted into the optimizing function \(A\))

\(A(x)=2x(12-x^2)=24x-2x^3\)

The domain here is that \(x\) must put us into the first quadrant, so \(x\in [0,2\sqrt{3}]\)

Differentiating gives us \(A'(x)=24-6x^2\), which always exists.

So our only critical points can be when \(A'(x)=0 \longrightarrow x=\pm 2\)

Since our domain \([0,2\sqrt{3}]\) we can ignore \(x=-2\) and just plug in the remaining critical point and domain endpoints to find our maximum for \(A(x)\):

\(A(0)=24\cdot 0-2\cdot 0^3=0\)
\(A(2)=24\cdot 2-2\cdot 2^3=32\)
\(A(2\sqrt{3})=24\cdot 2\sqrt{3}-2\cdot (2\sqrt{3})^3=0\)

Therefore our maximum possible area of the rectangle is 32
(there no no units provided, so we don't need to use them in our answer)