Optimization Optimization

Introduction

We can use local minima and maxima to optimize functions.

In solving optimization problems, we will use the same steps we use in other application problems:
1. Understand what the question is asking.
2. Convert the request into mathematical notation.
3. Solve the differential equation(s).
4. Interpret the solution.

Fencing

A farmer has 100 meters of fencing and wants to enclose a rectangular plot of land next to a river. The farmer doesn't need fencing along the river.

1. We want to maximize the area enclosed by the fence and river.

2. With

A
as the area,
l
as the length of the fence opposite the river, and
w
as the width of the fence adjacent to the river, we have
A = lw
. We also have
2w + l = 100
. Thus,
A = w(100 - 2w)
, on the interval
0 \lt w \lt 50
.

3. We want to find the local maxima of the function, which is where the derivative is
0
.
\frac{dA}{dw} = \frac{d}{dw}100w - 2w^2 = 100 - 4w
w = 25
4. The area inside the fencing is at its greatest when the width is 25 meters.
The farmer should make a fence with a width of 25 meters and a length of 50 meters.
The Area will be
1250 \, m^2
.

Shortest Distance

A woman is walking on the path

y = x^2
. She is starting at
(-10,100)
and ending at
(10,100)
. There is a hot dog cart at
(5,4)
. She wants to step off the path at the point closest to the cart, but she does not know which point that is.

1. We want to find the point on the path which is closest to the hot dog cart.

2. Find the point on
y=x^2
where the distance to
(5,4)
is minimized.
D
is the distance function between a point on the path and the hot dog cart.
D = \sqrt{(x-5)^2 + (y-4)^2}
.
We want the local minimum of D, where the derivative is
0
.

3.
\begin{align*} D &= \sqrt{(x-5)^2 + (y-4)^2} \\\\ &= \sqrt{(x-5)^2 + (\c1{x^2}-4)^2} \\\\ \frac{\c1{d}D}{\c1{dx}} &= \c1{\frac{d}{dx}} \sqrt{(x-5)^2 + (x^2-4)^2} \\\\ &= \frac{d}{dx} [(x-5)^2 + (x^2-4)^2]^{\c1{1/2}} \\\\ &= \c1{\frac{1}{2}}[(x-5)^2 + (x^2-4)^2]^{\c1{-1/2}} * \c1{\frac{d}{dx} [(x-5)^2 + (x^2-4)^2]} \\\\ &= \frac{1}{2}\frac{\c1{1}}{\sqrt{(x-5)^2 + (x^2-4)^2}} * [\c1{2}(x-5) + \c1{2}(x^2-4)\c1{2x}] \\\\ &= \frac{\c1{(x-5) + (x^2-4)2x}}{\sqrt{(x-5)^2 + (x^2-4)^2}} \\\\ \frac{dD}{dx} &= 0 \text{ when } [2(x-5) + 2(x^2-4)2x]=0 \\\\ \end{align*} \\ \begin{align*} 0 &= 2(x-5) + 2(x^2-4)2x \\\\ &= 2x - 10 + 4x^3 - 16x \\\\ &= 4x^3 \c1{- 14x} - 10 \\\\ &= (x+1)(2x^2-2x-5) \end{align*}
\frac{dD}{dx}=0
at
x=-1, \frac{1 \pm \sqrt{11}}{2}
.

Only
\frac{1 + \sqrt{11}}{2}
is in our domain, so
x=\frac{1 + \sqrt{11}}{2}
.

4. The point on the path closest to the hot dog cart is
(\frac{1 + \sqrt{11}}{2}, \left[\frac{1 + \sqrt{11}}{2}\right]^2)
.

Cost

A company's revenue function is

R(x) = 100x - 0.01x^2
and cost function is
C(x) = 50x + 2000
. How many units should the company produce to maximize profit?

1. We want to maximize the profit function,
P(x) = R(x) - C(x)
.

2.
P(x) = 100x - 0.01x^2 - 50x - 2000
We want to find the local maxima of the function, which is where the derivative is
0
.

3.
\begin{align*} P(x) &= -0.01x^2 + 100x - 50x - 2000 \\\\ &= -0.01x^2 + 50 - 2000 \\\\ \frac{d}{dx}P(x) &= \frac{d}{dx} -0.01x^2 + 50 - 2000 \\\\ &= -0.02x + 50 \\\\ x &= 2500 \end{align*}
4. The company should produce 2500 units to maximize profit. They will profit $62,500.