Extremum problems

An extremum problem looks for the greatest or the least value of a quantity – a profit, an area, a distance – written as a function of one variable.

• The variable: choose one quantity and call it $\displaystyle x$.
• The function: write the quantity you are looking for in terms of $\displaystyle x$ alone; write every other quantity in terms of $\displaystyle x$ first.
• The domain: the values $\displaystyle x$ can take – a positive length, a point on the graph, a price that is not negative.
• The derivative: differentiate and solve $\displaystyle {f'(x)}={0}$; keep only the solutions in the domain.
• The kind of extremum: from a table of the signs of $\displaystyle f'(x)$. On a closed domain, compare the values at the ends of the domain too.
• The answer: answer what was asked – sometimes $\displaystyle x$, sometimes the greatest value itself.

The sum of two positive numbers is 12.
We look for the numbers for which the square of the first times the second is the greatest.

The solution below calls the first number $\displaystyle x$, writes the second in terms of it, and finds the maximum of the product.

• The second number: $\displaystyle 12-x$, and the domain: $\displaystyle 0<x<12$.
• The product: $\displaystyle {f(x)}={x^2(12-x)}={12x^2-x^3}$.
• The derivative: $\displaystyle {f'(x)}={24x-3x^2}={3x(8-x)}$, which is 0 when $\displaystyle {x}={0}$ (not in the domain) or $\displaystyle {x}={8}$.

• The numbers: 8 and 4, and the greatest product: $\displaystyle {8^2\cdot4}={256}$.

A factory makes a product that costs 20 ₪ an item to make.
At a price of 40 ₪ an item the factory sells 100 items a day.
For every shekel taken off the price, the factory sells 10 more items a day.
$\displaystyle x$ – the amount taken off the price, in shekels.

The solution below writes the daily profit as a function of $\displaystyle x$ and finds its maximum.

• The price: $\displaystyle 40-x$, and the profit on each item: $\displaystyle {(40-x)-20}={20-x}$.
• The number of items a day: $\displaystyle 100+10x$.
• The daily profit: $\displaystyle {P(x)}={(20-x)(100+10x)}={2000+100x-10x^2}$, on the domain $\displaystyle 0\leq x\leq20$ (beyond 20 every item is sold at a loss).
• $\displaystyle {P'(x)}={100-20x}$, which is 0 when $\displaystyle {x}={5}$; to the left of 5 the derivative is positive, to the right negative – a maximum.
• At the ends of the domain: $\displaystyle {P(0)}={2000}$, $\displaystyle {P(20)}={0}$ – both less than $\displaystyle P(5)$.

• The factory should take 5 ₪ off the price, down to 35 ₪, and the greatest daily profit is 2,250 ₪.

The board shows the graphs of $\displaystyle {f(x)}={-x^2+9}$ and $\displaystyle {g(x)}={(x-3)^2}$.
The point $\displaystyle A$ is on the graph of $\displaystyle f$, in the first quadrant, between the points where the graphs meet.
A line through $\displaystyle A$ parallel to the $\displaystyle y$-axis meets the graph of $\displaystyle g$ at the point $\displaystyle B$.
A line through $\displaystyle A$ parallel to the $\displaystyle x$-axis meets the graph of $\displaystyle f$ at the point $\displaystyle C$.
$\displaystyle t$ – the $\displaystyle x$-coordinate of $\displaystyle A$.

The solution below writes the area of the triangle $\displaystyle ABC$ in terms of $\displaystyle t$, and finds the $\displaystyle t$ for which the area is the greatest. The board shows the case $\displaystyle {t}={2}$.

• The graphs meet where $\displaystyle {-x^2+9}={(x-3)^2}$, that is $\displaystyle {2x^2-6x}={0}$: at $\displaystyle {x}={0}$ and at $\displaystyle {x}={3}$. So $\displaystyle 0<t<3$.
• $\displaystyle {A}={(t,9-t^2)}$, $\displaystyle {B}={(t,(t-3)^2)}$, and $\displaystyle {C}={(-t,9-t^2)}$ – the point symmetric to $\displaystyle A$ about the $\displaystyle y$-axis.
• $\displaystyle {AB}={9-t^2-(t-3)^2}={6t-2t^2}$, and $\displaystyle {AC}={2t}$; the angle at $\displaystyle A$ is a right angle.
• The area: $\displaystyle {S(t)}={\frac{1}{2}\cdot2t\cdot(6t-2t^2)}={6t^2-2t^3}$.
• $\displaystyle {S'(t)}={12t-6t^2}={6t(2-t)}$, which is 0 inside the domain only when $\displaystyle {t}={2}$: positive to the left, negative to the right – a maximum.
• The greatest area: $\displaystyle {S(2)}={24-16}={8}$.

In the rectangle $\displaystyle ABCD$: $\displaystyle {AB}={18}$ cm, $\displaystyle {BC}={10}$ cm.
The parallelogram $\displaystyle PQTE$ is inscribed in the rectangle: $\displaystyle P$ on $\displaystyle AD$, $\displaystyle Q$ on $\displaystyle AB$, $\displaystyle T$ on $\displaystyle BC$ and $\displaystyle E$ on $\displaystyle CD$.
$\displaystyle {AP}={TC}$, and $\displaystyle {AQ}={EC}={3AP}$.

The solution below puts $\displaystyle {AP}={x}$, writes the area of the parallelogram in terms of $\displaystyle x$, and finds the $\displaystyle x$ for which the area is the greatest.

• $\displaystyle {AQ}={EC}={3x}$; for $\displaystyle Q$ to be on $\displaystyle AB$ we need $\displaystyle 3x\leq18$, so $\displaystyle 0<x\leq6$.
• The area of the parallelogram is the area of the rectangle less four right-angled triangles. The triangles at the corners $\displaystyle A$ and $\displaystyle C$ are congruent, and so are those at $\displaystyle B$ and $\displaystyle D$.
• At the corner $\displaystyle A$: the legs $\displaystyle x$ and $\displaystyle 3x$, the area $\displaystyle \frac{3x^2}{2}$. At the corner $\displaystyle B$: the legs $\displaystyle 18-3x$ and $\displaystyle 10-x$.

• $\displaystyle {S'(x)}={48-12x}$, which is 0 when $\displaystyle {x}={4}$, inside the domain: positive to the left, negative to the right – a maximum.
• $\displaystyle {AP}={4}$ cm, and the greatest area: $\displaystyle {S(4)}={192-96}={96}$ cm².

The sum of two numbers is 10.
$\displaystyle x$ – the first number.
We look for the numbers for which the sum of their squares is the least.

Differentiate the sum of the squares.

Solve the equation: the derivative equals 0.

What is the least sum of the squares?

A ticket for a show costs 60 ₪, and 400 people come to every show.
For every shekel added to the price, 4 fewer people come.
$\displaystyle x$ – the amount added to the price, in shekels.

Write the takings from one show as a function of $\displaystyle x$.
What is the domain of $\displaystyle x$?

Write an equation and solve it: what rise gives the greatest takings?

The theatre decides to raise the price by 15 ₪ at most.
At what price are the takings now the greatest, and what are they?
Give a reason from the derivative on the domain $\displaystyle 0\leq x\leq15$.

The board shows the graph of $\displaystyle {f(x)}={12-x^2}$.
A rectangle has two vertices on the $\displaystyle x$-axis and two vertices on the graph, in the first and in the second quadrant.
$\displaystyle x$ – the $\displaystyle x$-coordinate of the vertex on the graph in the first quadrant.

Show that the area of the rectangle is $\displaystyle {S(x)}={24x-2x^3}$.
What is the domain of $\displaystyle x$?

Find $\displaystyle S'(x)$.

Solve the equation $\displaystyle {S'(x)}={0}$.

What is the greatest area of the rectangle?

The vertex $\displaystyle B$ of the rectangle $\displaystyle ABCD$ is on the graph of $\displaystyle {f(x)}={\sqrt{2x-4}}$.
The side $\displaystyle AD$ is on the line $\displaystyle {x}={10}$, and the side $\displaystyle DC$ on the $\displaystyle x$-axis.
The graph of $\displaystyle f$ meets the $\displaystyle x$-axis at the point $\displaystyle F$.
$\displaystyle x$ – the $\displaystyle x$-coordinate of $\displaystyle B$; the board shows the case $\displaystyle {x}={4}$.

Write the area of the rectangle as a function of $\displaystyle x$.
What is the domain of $\displaystyle x$?

Differentiate the area.

Solve the equation: the derivative equals 0.
What are the coordinates of $\displaystyle B$ when the area of the rectangle is the greatest?

Find the area of the triangle $\displaystyle BFC$ when the area of the rectangle is the greatest.

The point $\displaystyle P$ is on the graph of $\displaystyle {f(x)}={\sqrt{x}}$.
We look for the point $\displaystyle P$ nearest to the point $\displaystyle (4,0)$.
$\displaystyle x$ – the $\displaystyle x$-coordinate of $\displaystyle P$.

Show that the distance from $\displaystyle P$ to $\displaystyle (4,0)$ is $\displaystyle {d(x)}={\sqrt{x^2-7x+16}}$.

Find $\displaystyle d'(x)$.

Solve the equation $\displaystyle {d'(x)}={0}$.

What are the coordinates of $\displaystyle P$, and what is the least distance?