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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=600−2lw=600-2l.

Substitute this into the area equation:

A=l(600−2l)A=l\left(600-2l\right)

A=600l−2l2A=600l-2l^2


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

A′=600−4lA'=600-4l and A′′=−4A''=-4


To find the critical points, we set the first derivative = 0:

600−4l=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,x≥0y=\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=(x−x0)2+(y−y0)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=(x−x0)2+(y−y0)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=(x−x0)2+(x−y0)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=(x−3)2+(x−0)2d^2 = (x-3)^2 + (\sqrt x-0)^2


=(x−3)2+x=(x-3)^2+x


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


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


⇒0=2(x−3)+1\Rightarrow0=2(x-3)+1


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


⇒x=52→critical 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



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Find the area of the largest rectangle that has two vertices on the xx-axis and two vertices lying on the graph of y=8−x2y=8-x^2 with

−8≤x≤8-\sqrt{8}\leq x\leq \sqrt{8}.




Choose an answer

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Extra Practice