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Optimization

The technique of finding a function's maximum and minimums can be applied to Optimization problems. These problems involve find the optimal value of a function, usually with a constraint equation.

Procedure for Optimization Problems

  1. Determine what you are minimizing/maximizing and draw a picture if possible
  2. Write an equation for the quantity you are trying to maximize or minimize (If there is more than 1 independent variable, create another equation relating the variables)
  3. Combine these to get an equation with an independent and a dependent variable
  4. Find and classify the critical points of this final combined equation
  5. Answer the question
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Example: Optimization

A farmer wishes to create a rectangular fenced enclosure for his animals. He has 600 ft. of fencing material and will construct the enclosure using a pre-existing fence as one side. Find the dimensions of the fenced area to maximize the enclosure.

1. Determine what you are minimizing/maximizing and draw a picture if possible
We want to MAXIMIZE the area of the rectangle

2. Write an equation for the quantity you are trying to maximize or minimize
*If there is more than 1 independent variable, create another equation relating the variables
The equation for area is A=l×wA=l\times w

We have 2 independent variables (length and width), so we need another equation that relates these 2 variables. Since the fence only has to make up 3 of the 4 sides of the rectangle, the equation is: 2l+w=6002l+w=600

3. Combine these to get an equation with an independent and a dependent variable
From the second equation, we have w=6002lw=600-2l.
Substitute this into the area equation:
A=l(6002l)A=l\left(600-2l\right)
A=600l2l2A=600l-2l^2

4. Find and classify the critical points of this final combined equation
A=6004lA'=600-4l and A=4A''=-4

To find the critical points, we set the first derivative = 0:
6004l=0      l=150600-4l=0\ \ \ \to\ \ \ l=150

Check to make sure this is a maximum:
A(150)=4<0A''\left(150\right)=-4<0
So, this critical point corresponds to a maximum.

5. Answer the question
Therefore, the dimensions that will maximize the enclosure are l=150 ftl=150\ ft and w=300 ftw=300\ ft.
The maximum area is 150×300=45000 ft2150\times300=45000\ ft^2.
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Example: Optimization

Find the point on the graph of the equation y=x,x0y=\sqrt{x}, x\geq 0, that is the closest to the point (3,0)(3,0).

1. Determine what you are minimizing/maximizing and draw a picture if possible

We are trying to minimize the distance between y=xy=\sqrt x and (3,0)(3,0).
2. Write an equation for the quantity you are trying to maximize or minimize
*If there is more than 1 independent variable, create another equation relating the variables

We want to optimize the distance formula d=(xx0)2+(yy0)2d = \sqrt{(x-x_0)^2 + (y-y_0)^2 }

However, for purposes of the derivative, it is equivalent and easier to use the square of the distance formula.

d2=(xx0)2+(yy0)2d^2 = (x-x_0)^2 + (y-y_0)^2

Since this equation has more than 1 independent variable, substitute the constraint equation y=xy=\sqrt x

3. Combine these to get an equation with an independent and a dependent variable

d2=(xx0)2+(xy0)2d^2 = (x-x_0)^2 + (\sqrt x-y_0)^2

Also we may substitute (3,0)(3,0) for x0x_0 and y0y_0

d2=(x3)2+(x0)2d^2 = (x-3)^2 + (\sqrt x-0)^2

=(x3)2+x=(x-3)^2+x

4. Find and classify the critical points of this final combined equation

ddx[d2]=ddx(x3)2+x\displaystyle \frac{d}{dx}[d^2]=\frac{d}{dx}(x-3)^2+x

0=2(x3)+1\Rightarrow0=2(x-3)+1

x3=12\displaystyle\Rightarrow x-3=-\frac{1}{2}

x=52critical point\displaystyle \Rightarrow x=\frac{5}{2} \rightarrow \text{critical point}

f(x)=2>0 local, min, at x=52\displaystyle f''(x) = 2 > 0 \therefore \text{ local, min, at } x= \frac{5}{2}

5. Answer the question

Substitute x=52\displaystyle x=\frac{5}{2} into y=x    y=52\displaystyle y=\sqrt x \implies y=\sqrt{\frac{5}{2}}

So the point (52,52)\left(\dfrac{5}{2},\sqrt{\dfrac{5}{2}}\right) on x\sqrt{x} is closest to (3,0)(3,0)

Four square corners are cut from a rectangular sheet measuring 160 cm by 160cm, and the remaining shape is bent to form an open-topped box. Find the side length of the cut-out square that maximizes the volume of the box.

Enter your answer in cm, do NOT include units


Find the area of the largest rectangle that has two vertices on the xx-axis and two vertices lying on the graph of y=8x2y=8-x^2 with
8x8-\sqrt{8}\leq x\leq \sqrt{8}.



Extra Practice