Optimisation

Q1

Given a fence of length $500m$, we want to enclose a rectangular area that is as big as possible. One side of the field is blocked by the farm house, so no fence is required for this side (see figure). Determine the dimension of the rectangle.

Q2

Consider again the fence from above. Find the smallest possible fence length (three sides) to enclose an area of $80\,000m^2$.

Q3

We want to construct a box whose base length is $3$ times the base width. The material used to build the top and bottom costs $10$ dollars per $m^2$ and the material used to build the sides costs $6$ dollars per $m^2$. If the box must have a volume of $50 m^3$, determine the dimensions that will minimize the cost to build the box.

Q4

Coke Can Problem. A manufacturer needs to make a cylindrical can that will hold $0.33$ litres of coke. Determine the optimal radius of the can that will minimize the amount of material used in its construction.

Q5

Find the point $Z$ on the graph of $f(x)=\sqrt{x}$ that is nearest to the point $A(4|0)$.

Q6

There are $50$ apple trees in an orchard. Each tree produces $n=800$ apples. For each additional tree planted in the orchard, the output per tree drops by $10$ apples. Find the optimal number of trees that should be added to the existing orchard in order to maximize the total output of trees?

Q7

Which cuboid with a square base and a volume of $1000 m^3$ has a minimal surface area?

Q8

A rectangular area $A$ is to be marked out with a rope of length $100m$. Part of one rectangle side is formed by a $20m$ long wall of a house, for which no rope has to be used (see picture). Find the largest possible area $A$ that can be formed.

Q9

The point $P$ is on the graph of the function $f(x)=-x+6$ in the first quadrant (that is, in the part of the coordinate system where the $x$- and $y$-coordinates are both positive). Find $P$ so that the rectangular area between the coordinate origin and $P$ is as large as possible.

Q10

A rectangle has two points on the graph of $f(x)=2-x^2$ and two points on the $x$-axis (see picture below). Determine the rectangle with the maximum area that can be embedded between the $x$-axis and the graph in this way (and is above the $x$-axis). How big is the area?

Solution

A7

Volume:

$$x^2y = 1000 \rightarrow y=\frac{1000}{x^2}$$

Cost function:

$$S(x)=2x^2+4xy=2x^2+4x\frac{1000}{x^2}=2x^2+4000x^{-1}$$

Find local min:

$$S'(x)=4x-4000x^{-2}=0$$

Multiplying both sides by $x^2$ we get

$$4x^3-4000=0 \rightarrow x=10m$$

And therefore $y=\frac{1000}{10^2}=10m$. Thus, the solution is therefore a cube of side length $10m$.

Extra: check if local minimum.

$$S''(x)=4+8000x^{-3} \rightarrow A''(10)=4+8=12>0\rightarrow \text{ local min.}$$

A8

Rope length:

$$x+2y+(x-20)=100 \rightarrow y=60-x$$

Cost function:

$$A(x)=xy=x(60-x)=60x-x^2$$

Find local max:

$$A'(x)=60-2x\rightarrow x=30m$$

And thus $y=60-30=30m$ and the maximal area is $A=900m^2$.

Extra: check if local maximum.

$$A''(x)=-2 \rightarrow A''(30)=-2\rightarrow \text{ local max.}$$

A9

The condition is that $P$ hs to be on the graph $f(x)=-x+6$, that is, the $y$-coordinate of $P$ is given by

$$y=f(x)=-x+6$$

Cost function:

$$A(x)=xy=x(-x+6)=-x^2+6x$$

Find local max:

$$A'(x)=-2x+6=0 \rightarrow x=3$$

Thus $y=-3+6=3$ and therefore $P(3|3)$.

Extra: check if local maximum.

$$A''(x)=-2 \rightarrow A''(3)=-2\rightarrow \text{ local max.}$$

A10

The width of the rectangle is $2x$, the height is $f(x)=2-x^2$. The area is therefore $A(x)=2x\cdot (2-x^2)$. We have to find the local maximum of this cost function. First let's expand $A(x)$: $A(x)=4x-2x^3$, thus

$A'(x)= 4-6x^2=0 \rightarrow x=\sqrt{\frac{2}{3}}$

Is it really a maximum? Yes, because

$A''(x)=-12x$ thus $A''(\sqrt{\frac{2}{3}})=-12\cdot\sqrt{\frac{2}{3}} <0 \rightarrow$ local max!

The width of the rectangle is $2\sqrt{\frac{2}{3}}=1.633$, the height is $2-2\cdot\frac{2}{3}=\frac{2}{3}$.