This lesson offers a variety of word/story problems
that are solved by finding the maximum or minimum value of a
quadratic function.
Suggested review:
General Strategy for Optimization (Max/Min) Problems

[What are you maximizing/minimizing?]
Read the problem carefully.
Answer the question: ‘What am I being asked to maximize/minimize?’
Look for ‘extreme’ words—for example: maximum,
minimum, greatest, least, biggest, smallest, largest, most.

[write expression]
Write an expression for the thing you need to maximize/minimize (from step 1).
You may want to make a sketch.
You may need to give names to unknown quantities.

[one variable only]
Does your expression from step 2 involve more than one variable?
If so, look for other known information that allows you to eliminate variable(s).
You must end up with an expression that involves only one variable.
Give a meaningful function name to your expression, so function notation can be
used to write your work concisely.

[find and report]
Use appropriate techniques to find the desired maximum/minimum value, and where it occurs.
Use a complete sentence to report your result.
Example (Max/Min problem)
Among all rectangles that have a perimeter of twenty feet, find the dimensions of
the one with the largest area.
Presolution: Stop and think! Do the dimensions actually affect the area?
$1\text{ ft} \times 9\text{ ft}$ rectangle: perimeter is $\,20\text{ ft}\,$, area is $\,9\text{ sq ft}$
$2\text{ ft} \times 8\text{ ft}$ rectangle: perimeter is $\,20\text{ ft}\,$, area is $\,16\text{ sq ft}$
Yes, the dimensions do affect the area.
(You've probably already guessed the answer at this point!
But, this is a nice simple question to clearly illustrate the general procedure.)
SOLUTION:
 [What are you maximizing/minimizing?]
key word: ‘largest’
Largest what? Largest area.
You need to maximize the area of a rectangle.
 [write expression]
You need an expression for the area of a rectangle: area equals length times width.
Choose meaningful names: let $\,\ell\,$ be the length, $\,w\,$ be the width, and $\,A\,$ be the area.
Let $\,w\,$ and $\,\ell\,$ have units of feet; $\,A\,$ therefore has units of square feet.
With these names: $A = \ell w$
At this moment, the expression for $\,A\,$ involves two variables, not one.

[one variable only]
Reread the problem.
Can we write down anything that is true that involves our named variables?
Yes—the rectangle must have a perimeter of $\,20\,$ feet.
Thus, $\,2\ell + 2w = 20\,$.
This allows us to solve for one variable in terms of the other.
You can solve for $\,\ell\,$ in terms of $\,w\,$, or vice versa—your choice.
Sometimes it's easier to solve for one variable than another, but here there's no difference.
I choose to solve for $\,\ell\,$:
$2\ell + 2w = 20$
$\ell + w = 10$
$\ell = 10w \ \ \ \ (*)$
So: $A = \ell w = (10w)w$
At this point, $\,A\,$ is a function of only one variable, $\,w\,$.
Switch to function notation:
$A(w) = (10w)w = 10w  w^2 = w^2 + 10w$
Note that $\,A\,$ is a quadratic function that is concave down (sheds water), so it has a maximum value.

[find and report]
The vertex formula says that the vertex of $\,f(x) = ax^2 + bx + c\,$ is
$\,\bigl(\frac{b}{2a},f(\frac{b}{2a})\bigr)\,$.
For $\,A(w) = w^2 + 10w\,$, we have $\,a = 1\,$, $\,b = 10\,$.
Thus, $\displaystyle\,\frac{b}{2a} = \frac{10}{2(1)} = 5\,$; this is the $\,x\,$value of the vertex of the quadratic area function.
That is, when the width is $\,5\,$ ($\,w = 5\,$) the maximum area occurs.
Note from (*) that when $\,w = 5\,$, we have $\,\ell = 10w = 105 = 5\,$.
The rectangle that achieves the maximum area is actually a $\,5\times 5\,$ square.
What is the maximum area?
$A(5) = 5^2 + 10\cdot 5 = 25 + 50 = 25\,$; this is the $\,y\,$value of the vertex of the quadratic area function.
Among all rectangles with a perimeter of $\,20\,$ feet, the one with
dimensions $\,5\text{ ft}\times 5\text{ ft}\,$ has the largest area;
this largest area is $\,25\,$ square feet.
Example (Max/Min problem)
Find two positive numbers $\,x\,$ and $\,y\,$ that add to $\,30\,$, and for
which $\,2x^2 + 5y^2\,$ is a minimum.
Presolution: Stop and think! Get a feeling for the problem.
$x = 1\,$, $\,y = 29\,$: $2x^2 + 5y^2 = 2(1^2) + 5(29^2) = 4207$
$x = 2\,$, $\,y = 28\,$: $2x^2 + 5y^2 = 2(2^2) + 5(28^2) = 3928$
Yes, the choices for $\,x\,$ and $\,y\,$ do affect the quantity $\,2x^2 + 5y^2\,$.
For this question, you're probably not able to guess the answer.
There's no restriction that $\,x\,$ and $\,y\,$ must be whole numbers, so
$\,x = 1.5\,$ and $\,y = 28.5\,$ is allowable.
SOLUTION:
 [What are you maximizing/minimizing?]
key word: ‘minimum’
You need to minimize the quantity $\,2x^2 + 5y^2\,$.
 [write expression]
We already have the expression; let's call it $\,S\,$ for sum.
So, $\,S = 2x^2 + 5y^2\,$.
At this moment, $\,S\,$ is a function of two variables.

[one variable only]
Reread the problem.
Can we write down anything that is true that involves $\,x\,$ and $\,y\,$?
Yes—they must add to $\,30\,$.
Thus, $\,x + y = 30\,$.
This allows us to solve for one variable in terms of the other.
Choosing to solve for $\,y\,$ gives:
$y = 30  x\ \ \ (*)$
So: $S = 2x^2 + 5y^2 = 2x^2 + 5(30x)^2 = \text{(you check these steps)} = 7x^2  300x + 4500$
At this point, $\,S\,$ is a function of only one variable, $\,x\,$.
Switch to function notation:
$S(x) = 7x^2  300x + 4500$
Note that $\,S\,$ is a quadratic function that is concave up (holds water), so it has a minimum value.

[find and report]
The vertex formula says that the vertex of $\,f(x) = ax^2 + bx + c\,$ is
$\,\bigl(\frac{b}{2a},f(\frac{b}{2a})\bigr)\,$.
For $\,S(x) = 7x^2  300x + 4500\,$, we have $\,a = 7\,$, $\,b = 300\,$.
Thus, $\displaystyle\,\frac{b}{2a} = \frac{(300)}{2(7)} = \frac{150}{7}\,$; this is the $\,x\,$value of the vertex of the quadratic function $\,S\,$.
That is, when $\,x = \frac{150}{7}$ the minimum value occurs.
Note from (*) that when $\,x = \frac{150}{7}\,$, we have $\,y = 30  \frac{150}{7} = \frac{210}{7}  \frac{150}{7} = \frac{60}{7}\,$.
What is the minimum value of $\,2x^2 + 5y^2\,$?
$S(\frac{150}{7}) = 2(\frac{150}{7})^2 + 5(\frac{60}{7})^2 = \text{(after a bit of work)} = \frac{9000}{7}\,$; this is the $\,y\,$value of the vertex of the quadratic function $\,S\,$.
The two positive numbers $\,x\,$ and $\,y\,$ that add to $\,30\,$ and make $2x^2 + 5y^2\,$ as
small as possible are $\,x = \frac{150}{7}\,$ and $\,y = \frac{60}{7}\,$;
this smallest value of $\,2x^2 + 5y^2\,$ is $\,\frac{9000}{7}\,$.
To make your life easier:
zip up to WolframAlpha and type in:
vertex of 2x^2 + 5(30x)^2
How easy is that?!