Optimization Problems
Ximena wants to build a pen for her rabbits using a fixed amount of fencing. What shape should it have so the space is as large as possible? And a can factory asks: which dimensions use the least amount of sheet metal? All these questions look for an optimal value (the greatest or the least) of a quantity that depends on a measurement we can choose. The derivative is the tool for finding it.
The general procedure, step by step
Split the number $24$ into the sum of two positive numbers such that their product is maximum. Set up the function to be optimised and solve it using differentiation.
Split the number $24$ into the sum of two positive numbers such that their product is maximum. Set up the function to be optimised and solve it using differentiation.
Which positive number exceeds its square by the greatest possible amount? Express this amount as a function of the number and maximise it using differentiation.
Which positive number exceeds its square by the greatest possible amount? Express this amount as a function of the number and maximise it using differentiation.
Find two positive numbers that sum to $18$ and whose sum of squares is minimum.
Find two positive numbers that sum to $18$ and whose sum of squares is minimum.
Geometric problems: areas and perimeters
These are the most common: there is a figure (rectangle, pen, box) with a fixed measurement (perimeter, amount of fencing, sheet metal) and you are asked for the shape that makes another quantity (area, volume) a maximum or a minimum.
Among all rectangles with a perimeter of $36\ \text{m}$, find the dimensions of the one with maximum area and calculate that area.
Among all rectangles with a perimeter of $36\ \text{m}$, find the dimensions of the one with maximum area and calculate that area.
A gardener wants to fence off a rectangular vegetable patch against a wall of his house, so that he only needs to fence three sides. He has $60\ \text{m}$ of fencing…
A gardener wants to fence off a rectangular vegetable patch against a wall of his house, so that he only needs to fence three sides. He has $60\ \text{m}$ of fencing available.
- a) Express the area of the vegetable patch as a function of the side $x$ perpendicular to the wall.
- b) Find the dimensions that maximise the area and the value of that area.
An open-topped box is to be made from a square piece of cardboard with sides of $24\ \text{cm}$ by cutting out four equal squares of side length $x$ from the corners and…
An open-topped box is to be made from a square piece of cardboard with sides of $24\ \text{cm}$ by cutting out four equal squares of side length $x$ from the corners and folding up the sides, as shown in the figure.
- a) Express the volume of the box $V(x)$ as a function of $x$ and state the valid domain for $x$.
- b) Find the value of $x$ that maximises the volume.
- c) Calculate the maximum volume and check, using the sign of $V'(x)$ on both sides, that it is indeed a maximum.
Real-life problems: costs, boxes, and models
The same procedure works to minimize a cost, maximize the volume of a box with a given amount of material, or find the extreme of a model that depends on time or price.
A farmer has $240\ \text{m}$ of fencing to build a rectangular pen divided into two plots by an internal fence parallel to one of the sides, as shown in the figure. a)…
A farmer has $240\ \text{m}$ of fencing to build a rectangular pen divided into two plots by an internal fence parallel to one of the sides, as shown in the figure.
- a) Express the total area of the pen as a function of one of its sides.
- b) Determine the dimensions that maximise the total area and calculate that area.
A clothing store sells $80$ T-shirts each week for $\$12$ each. It estimates that for every pesos it reduces the price, it will sell $10…
A clothing store sells $80$ T-shirts each week for $\$12$ each. It estimates that for every pesos it reduces the price, it will sell $10$ more T-shirts per week.
- a) Express the weekly revenue $I(x)$ as a function of the price reduction $x$ in pesos.
- b) Find the price that maximizes the revenue and the maximum weekly revenue.
A beam of rectangular cross-section is to be cut from a log of circular cross-section of diameter $d$, as shown in the figure. The bending strength of the beam is…
A beam of rectangular cross-section is to be cut from a log of circular cross-section of diameter $d$, as shown in the figure. The bending strength of the beam is proportional to the product of its base and the square of its height: $R = k \times b \times h^2$.
- a) Express the strength as a function of the base $b$ using the relationship connecting $b$, $h$ and $d$.
- b) Find the dimensions of the beam with maximum strength.
Exploring a parameter and justifying an optimum
A rectangle with perimeter $p$ has area $A(x)=x(p/2-x)$, with $0 Reason whether the following statement is true or false: "The maximum value of a continuous function on a closed interval is always reached at a point where $f'(x) = 0$." If false, give a counterexample using a simple function on a specific interval. For the problem of the open box constructed from a square piece of card of side $32\ \text{cm}$, whose volume is $V(x) = x \times (32 - 2x)^2$ with $0 < x < 16$, a student writes: "$V'(x) = 12x^2 - 256x + 1024 = 0$ gives $x = 16$ and $x = \frac{16}{3}$; since volume increases with the size of the cut-out, I choose $x = 16\ \text{cm}$". Point out the two mistakes made in the conclusion (one concerning domain and one concerning justification) and solve the problem correctly. Back to the beginning: Ximena's pen with fixed fencing has maximum area when it is a square (or, next to a wall, when the side parallel to the wall is twice the other); and the can that uses the least sheet metal comes from applying the five steps with the volume constraint.Reason whether the following statement is true or false: "The maximum value of a continuous function on a closed interval is always reached at a point where…
For the problem of the open box constructed from a square piece of card of side $32\ \text{cm}$, whose volume is $V(x) = x \times (32 - 2x)^2$ with $0 < x < 16$, a…