MCV4U Optimization Problems: Notes and Practice

Optimization Problems

Optimization means finding the largest or smallest possible value of a quantity, such as the biggest box or the can that uses the least metal. Model the quantity as a function of one variable, use the context to fix the domain, then let the derivative find the best value.

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Key things to remember

1On a closed interval, the absolute maximum and minimum occur at critical numbers or endpoints: evaluate them all and compare.

2Use the constraint to write the quantity to optimize as a function of ONE variable, then state its domain.

3Reject critical numbers outside the domain; confirm a max or min with the first or second derivative test.

4Open box from an

L \times W

sheet:

V (x) = x (L - 2 x) (W - 2 x)

with

0 < x < \frac{min ( L , W )}{2}

5To minimize a distance, minimize

d^{2}

instead; the same

x

gives the minimum.

A packaging company wants a box that holds as much as possible from a fixed sheet of cardboard. A farmer wants to fence the biggest field with the fencing she owns. A drink company wants a can that holds 500 mL using as little aluminium as possible. Each is a trade-off: making one dimension bigger forces another to be smaller, so there is a best value somewhere in between. In Unit 3 you learned that, inside its domain, a smooth function can only reach a local maximum or minimum where its derivative is zero. Optimization turns a word problem into a function so that fact can find the best design.

The goal is the ABSOLUTE (global) maximum or minimum on the domain the situation allows. On a closed interval $a \leq x \leq b$ , a continuous function always has one, and it occurs either at a critical number inside the interval or at an endpoint, so you compare the values at all of them. On an open interval, check what happens to the function near the ends and use the first or second derivative test to confirm the critical number gives the extreme you want.

Fig 1The six steps of an optimization problem, with the river-fence example worked beside each one: draw and label, write the quantity as a function of one variable using the constraint, state the domain, differentiate and solve f′ = 0, test the critical points and endpoints, and answer with units.

Draw a diagram and label every quantity that can change. Name the quantity to optimize.
Write that quantity as a formula. If it has two variables, use the constraint (the fixed amount of fencing, cardboard or volume) to eliminate one.
State the domain from the context: lengths must be positive, a cut cannot exceed half the sheet, and so on.
Differentiate and solve $f^{'} (x) = 0$ . Reject any solution outside the domain.
Test: compare values at critical numbers and endpoints, or use the first or second derivative test.
Answer the question that was asked, in a sentence, with units. Often it asks for dimensions, not just the maximum value.

VideoOntario teacher JensenMath runs the same steps on a simple rectangle with a fixed perimeter: write the area as a function of one variable, find the critical number and confirm it gives the maximum.Watch on YouTube

The open box

Fig 2A 30 cm by 30 cm square sheet with an x by x square cut from each corner, folded into an open box. The small graph of V(x) = x(30 − 2x)² peaks at x = 5, where V = 2000 cm³.

Cut squares of side $x$ from the corners of a $30$ cm by $30$ cm sheet and fold up the sides. The base becomes $(30 - 2 x)$ by $(30 - 2 x)$ and the height is $x$ , so $V (x) = x (30 - 2 x)^{2}$ . Tiny squares give a flat box with almost no volume; squares close to $15$ cm leave almost no base. The domain is $0 < x < 15$ , and the best cut is somewhere in between. The product and chain rules give $V^{'} (x) = (30 - 2 x)^{2} - 4 x (30 - 2 x) = (30 - 2 x) (30 - 6 x),$ which is zero at $x = 15$ (outside the domain) and $x = 5$ . So the best box has height $5$ cm, base $20$ cm by $20$ cm, and volume $2000 cm^{3}$ .

VideoThe open box problem with a square sheet of cardboard and squares cut from the corners, solved with the derivative just like the example above.Watch on YouTube

Fencing beside a river

Fig 3A rectangular field along a straight river with 600 m of fence on the other three sides. A(x) = x(600 − 2x) is greatest when each side perpendicular to the river is x = 150 m, making the side along the river 300 m and the area 45 000 m².

With $600$ m of fence and no fence needed along the river, let $x$ be each side perpendicular to the river. The side parallel to the river gets what is left, $600 - 2 x$ , so $A (x) = x (600 - 2 x) = 600 x - 2 x^{2}$ for $0 < x < 300$ . Then $A^{'} (x) = 600 - 4 x = 0$ at $x = 150$ . $A^{'}$ is positive before $150$ and negative after, so this is the maximum: a $150$ m by $300$ m field of area $45000 m^{2}$ . Notice the answer is not a square, because the river saves one side of fencing.

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Choose the open box, the fence by a river or the can of least metal. Drag the slider to change the shape, watch the objective graph and the sign of the derivative, then press Find the optimum for the full calculus solution.

Always check the domain and the endpoints

Solving

f^{'} (x) = 0

only finds candidates. A root outside the domain (like

x = 15

for the box) must be rejected. On a closed interval, an endpoint can beat every critical number, so evaluate the function at the endpoints too.

Minimize the square of a distance

A distance

d = (x - a)^{2} + (y - b)^{2}

is smallest exactly when

d^{2}

is smallest, because the square root is an increasing function. Minimizing

d^{2}

avoids differentiating a square root and gives the same

x

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Squares of side

x

are cut from the corners of a

30

cm by

30

cm sheet, and the sides are folded up to make an open box. Find the maximum volume.

Example 2

A closed cylindrical can must hold

500 cm^{3}

. Find the radius and height that use the least metal.

Example 3

Find the point on the curve

y = x

that is closest to the point

(3, 0)

Example 4

A rectangle has its base on the

x

-axis and its upper corners on the parabola

y = 12 - x^{2}

. Find the largest possible area.

Where marks get lost

Leaving the function in two variables and differentiating anyway. Use the constraint to eliminate one variable first.

Keeping a critical number outside the domain, such as

x = 15

for the

30

cm box, which leaves no base at all.

Forgetting the endpoints of a closed interval: the absolute extreme can be at an endpoint, not at a critical number.

Stopping at

x = 5

when the question asks for the maximum volume or the dimensions. Reread the question and answer it in a sentence with units.

Assuming the best rectangle is always a square. With a river on one side, the best field is twice as long as it is wide.

Key things to remember

Worked examples

Where marks get lost

Flashcards 7

Practice