Application of Derivatives · Lesson 5 of 7

Building and Solving Optimization Models

“Build a one-variable objective from constraints and justify the greatest or least feasible outcome.”

Learning Objectives

• Choose variables, constraints, and feasible domains for optimization. • Build objectives for numbers, distances, areas, volumes, costs, and profit. • Use Pythagoras and similar triangles to eliminate extra variables. • Reject stationary roots that do not describe allowed configurations. • Justify a global optimum and report the original requested quantities.

From a Situation to a Function

Optimization means finding the greatest or least possible value while respecting given restrictions. The derivative is only one part of the work. First decide what may vary, what must remain fixed, and what output is being optimized. A correct derivative of the wrong model cannot solve the original problem.

Name the variable and state its feasible domain. Then use the constraint to remove extra variables. Once an objective function is written in one variable, find candidates, test them, and justify the comparison over the whole feasible set. Finish by translating the result back into the original quantities and units.

Stage	Question to ask
Variable	Which dimension, position, or output is free to vary?
Constraint	What stays fixed, and how are variables linked?
Objective	Which function represents the requested area, volume, distance, cost, or profit?
Feasible set	Which inputs describe actual allowed configurations?
Extremum check	Why is the candidate greatest or least over the allowed set?
Interpretation	What dimensions or value does the original problem request?

Numbers Subject to a Fixed Sum

A fixed total immediately connects two unknown numbers. Write the second in terms of the first, so that products or sums of powers become functions of one variable. Positivity matters: if both numbers must be positive, neither endpoint of the resulting interval represents an allowed pair.

Example — The least sum of squares

Problem
Find two positive numbers with sum 15 whose sum of squares is least.

1.Let the numbers be $x$ and $15 - x$ , with $0 < x < 15$ . The objective is $S (x) = x^{2} + (15 - x)^{2} = 2 x^{2} - 30 x + 225$ .
2.Differentiate: $S^{'} (x) = 4 x - 30$ , so the stationary candidate is $x = 15/2$ . Since $S^{''} (x) = 4 > 0$ , it is a local minimum.
3.To establish the whole-domain result, complete the square: $S (x) = 2 (x - 15/2)^{2} + 225/2$ . It is at least $225/2$ for every allowed input.
4.Thus both numbers are $15/2$ . More generally, for positive numbers with sum $k > 0$ , the least sum of squares occurs at $k /2, k /2$ .

The same setup can optimize a product. For a fixed sum $k$ , the product $x (k - x)$ has derivative $k - 2 x$ , positive before $k /2$ and negative after. Equal numbers give its greatest value. For weighted products such as $x^{2} y^{5}$ , keep the powers in the objective instead of assuming equality will still be optimal.

Minimizing a Distance or Its Square

Distance formulas often contain square roots. Because square root is strictly increasing on nonnegative numbers, the smallest distance occurs at exactly the same configuration as the smallest squared distance. Optimizing the square can simplify differentiation, but the final requested distance still needs a square root.

Example — Nearest point on a parabola

Problem
Find the shortest distance from $(0, c)$ to $y = x^{2}$ , where $1/2 \leq c \leq 5$ .

1.Write a point on the parabola as $(h, k)$ with $k = h^{2} \geq 0$ . Its squared distance is $G (k) = k + (k - c)^{2}$ . Every $k \geq 0$ corresponds to one or two points on the parabola.
2.Differentiate: $G^{'} (k) = 1 + 2 (k - c)$ , so the candidate is $k = c - 1/2$ . It is feasible because $c \geq 1/2$ .
3.Complete the square: $G (k) = (k - (c - 1/2))^{2} + c - 1/4$ . The first term is nonnegative, so the minimum over $k \geq 0$ is $c - 1/4$ .
4.The shortest distance is $c - 1/4 = 4 c - 1 /2$ . If $c > 1/2$ , the nearest points have $h = \pm c - 1/2$ and $k = c - 1/2$ .
5.At $c = 1/2$ , the optimum is the boundary $k = 0$ , giving the single nearest point $(0, 0)$ and distance $1/2$ . The boundary case must not be treated as an interior point.

Squared distances to two pole tops— R moves along the ground segment. Each slant distance is the hypotenuse of a right triangle.

Example — Choosing a ground position between poles

Problem
Two vertical poles have heights 16 m and 22 m, with bases 20 m apart. Find $R$ on the segment between the bases minimizing the sum of squared distances to the tops.

1.Let $A R = x$ , so $R B = 20 - x$ and $0 \leq x \leq 20$ . Pythagoras gives $R P^{2} = x^{2} + 1 6^{2}$ and $R Q^{2} = (20 - x)^{2} + 2 2^{2}$ .
2.The objective is $S (x) = 2 x^{2} - 40 x + 1140$ . Its derivative is $4 x - 40$ , zero at 10; its second derivative is 4.
3.Equivalently, $S (x) = 2 (x - 10)^{2} + 940$ , so 10 gives the absolute minimum over the allowed segment.
4.Therefore $A R = 10$ m. The unequal pole heights add constants to the squared-distance objective and do not shift the optimum. This conclusion concerns the sum of squared distances, not the sum of distances.

An Area Model Built from Geometry

In a geometric optimization problem, the main relationship often comes from a right triangle or similar triangles. A labelled diagram should display the variable and the fixed measurements. Express the height in terms of that variable before writing the area.

A trapezium with three sides of length 10— The horizontal offset x determines the height through a right triangle.

Example — The largest trapezium area

Problem
A trapezium has its top side and both equal sloping sides of length 10 cm. Find its greatest area.

1.For the outward-sloping configuration, let the horizontal offset of each side be $x$ , with $0 \leq x < 10$ . The height is $100 - x^{2}$ and the lower base is $10 + 2 x$ .
2.The area is $A (x) = (10 + x) 100 - x^{2}$ . Differentiating gives $A^{'} (x) = (100 - 10 x - 2 x^{2}) / 100 - x^{2}$ .
3.The numerator is $- 2 (x - 5) (x + 10)$ . Within the feasible interval, its sign is positive before 5 and negative after 5. The denominator is positive.
4.The area therefore rises to $x = 5$ and then falls, proving a greatest value there. It equals $1575 = 753$ cm².
5.An inward-sloping configuration has both height at most 10 and the other parallel side at most 10, so its area is at most 100 cm². It cannot improve on $753$ cm².

A Solid Inscribed in Another Solid

An inscribed cylinder cannot choose its radius and height independently. The boundary of the surrounding cone links them. Similar triangles in an axial cross-section turn that spatial restriction into a simple equation.

Cylinder inscribed in a cone— The cylinder is coaxial with the cone and shares its base plane. Similar triangles link its radius and height.

Example — Greatest curved surface area of a cylinder

Problem
A coaxial right circular cylinder is inscribed in a cone of radius $r$ and height $h$ . Find its radius when its curved surface area is greatest.

1.Let the cylinder radius be $x$ and its height $y$ . Similar triangles give $y / h = (r - x) / r$ , so $y = h (r - x) / r$ , with $0 < x < r$ .
2.Its curved surface area is $S (x) = 2 π x y = (2 π h / r) (r x - x^{2})$ . This is the curved area only, excluding the circular ends.
3.Differentiate: $S^{'} (x) = (2 π h / r) (r - 2 x)$ . It is positive before $r /2$ and negative after, so the greatest area occurs at $x = r /2$ .
4.The associated cylinder height is $h /2$ . The area tends to zero at the degenerate endpoint configurations, consistent with the interior maximum.

A Folded Box and a Feasible Root

A derivative equation may have several algebraic roots, but some may not describe a real object. For a box folded from a sheet, the remaining length and breadth must both be positive. Write these conditions before accepting a stationary point.

Cutting and folding an open box— Each cut has side x. The folded box has height x and base dimensions 8 − 2x and 3 − 2x.

Example — The largest open box

Problem
Equal squares are removed from the corners of a 3 m by 8 m sheet, then its sides are folded. Find the greatest box volume.

1.Let each cut have side $x$ . The dimensions are $x$ , $3 - 2 x$ , and $8 - 2 x$ , so $0 < x < 3/2$ .
2.The volume is $V (x) = x (3 - 2 x) (8 - 2 x) = 4 x^{3} - 22 x^{2} + 24 x$ .
3.Differentiate: $V^{'} (x) = 12 x^{2} - 44 x + 24 = 4 (x - 3) (3 x - 2)$ . The roots are 3 and $2/3$ , but 3 is outside the feasible domain.
4.On $(0, 2/3)$ the derivative is positive, and on $(2/3, 3/2)$ it is negative. Hence $2/3$ gives the greatest feasible volume.
5.The box dimensions are $2/3$ , $5/3$ , and $20/3$ metres. Their product is $200/27$ m³. The limiting volume at either endpoint is zero.

Keeping Surface Area and Volume Constraints Distinct

A curved surface excludes a solid’s circular base or ends; total surface area includes them when the solid is closed. An optimization result changes when those parts change. Write the exact area formula before eliminating a variable, especially in can and cone problems.

Solid and quantity	Formula	Constraint used to reduce variables
Closed cylinder surface area	$2 π r^{2} + 2 π r h$	Fixed volume gives $h = V / (π r^{2})$
Cone curved area	$π r r^{2} + h^{2}$	Fixed volume gives $h = 3 V / (π r^{2})$
Cone total area	$π r^{2} + π r r^{2} + h^{2}$	Include its base when total area is fixed
Cylinder inside a sphere of radius $R$	$V = π r^{2} h$	An axial right triangle gives $r^{2} + (h /2)^{2} = R^{2}$
Cone inside a sphere of radius $R$	$V = π r^{2} h /3$	Its base lies $R - h$ from the centre, so $r^{2} = R^{2} - (R - h)^{2}$

Example — A closed can using the least material

Problem
For a fixed volume $V_{0} > 0$ , find the relation between radius and height of a closed cylinder with least total surface area.

1.From $V_{0} = π r^{2} h$ , eliminate height: $h = V_{0} / (π r^{2})$ , with $r > 0$ .
2.The total area, including both ends, is $S (r) = 2 π r^{2} + 2 V_{0} / r$ . Differentiate: $S^{'} (r) = 4 π r - 2 V_{0} / r^{2}$ .
3.Multiplying the derivative by the positive $r^{2}$ shows its sign is the sign of $4 π r^{3} - 2 V_{0}$ . It changes from negative to positive at $r^{3} = V_{0} / (2 π)$ .
4.Thus this is the global minimum on $r > 0$ . Substituting $V_{0} = 2 π r^{3}$ into the height formula gives $h = 2 r$ .
5.For $V_{0} = 100$ cm³, the dimensions are $r = (50/ π)^{1/3}$ cm and $h = 2 (50/ π)^{1/3}$ cm. The least-material shape has height equal to diameter.

Profit Is Revenue Minus Cost

Selling more items increases the number sold but may require a lower unit price. Revenue is unit price multiplied by quantity, and profit subtracts total cost. Keep these three quantities separate while forming the model. For item counts, check that a proposed continuous optimum is also compatible with whole items.

Example — Production for greatest profit

Problem
A manufacturer sells $x$ items at unit price $5 - x /100$ rupees. Total cost is $x /5 + 500$ rupees. Find the quantity giving greatest profit.

1.Revenue is $R (x) = x (5 - x /100) = 5 x - x^{2} /100$ , not just the unit price.
2.Profit is $P (x) = R (x) - C (x) = 24 x /5 - x^{2} /100 - 500$ . Nonnegative quantity and price give a natural model domain $0 \leq x \leq 500$ .
3.Differentiate: $P^{'} (x) = 24/5 - x /50$ . It is zero at $x = 240$ . Also $P^{''} (x) = - 1/50 < 0$ .
4.The profit is a downward-opening quadratic and increases before 240 and decreases after. Thus 240 is the global maximizer in the model domain.
5.The quantity 240 is a whole number, so no rounding comparison is needed. The greatest profit is $P (240) = 1152 - 576 - 500 = 76$ rupees.

Example — The greatest height along a path

Problem
A path is modelled by $h (x) = 60 + x - x^{2} /60$ , where $x$ is horizontal distance. Find its greatest height while the object is above ground.

1.The derivative is $h^{'} (x) = 1 - x /30$ , zero at $x = 30$ . The second derivative is $- 1/30 < 0$ .
2.The path rises before 30 and falls afterwards. Its height there is $60 + 30 - 900/60 = 75$ .
3.The point $x = 30$ lies in the physical above-ground part of the path, so the greatest height is 75 in the given length units.

A profit model $P (x) = a x + b x^{2}$ has a similar shape only if $b < 0$ . Its stationary input is $- a / (2 b)$ , but this represents a physical maximizing quantity only if it lies in the permitted domain. If the coefficients or domain do not satisfy those conditions, a different comparison is needed.

A local check needs a global argument

For the greatest or least feasible value, justify why other inputs cannot do better. A sign change across the entire feasible interval, comparison with endpoints, or a complete-square expression can supply that argument. A local second derivative test alone is not a comparison with every feasible input.

Check Your Understanding

Use these questions to check both your calculations and your interpretation of the ideas. Read the conditions carefully before choosing an answer.

Quick check

Two positive numbers sum to $k$ . Which expression gives their sum of squares?

Quick check

Why can squared distance be minimized instead of distance?

Quick check

A box is cut from a 3 m by 8 m sheet. Which is its feasible cut interval?

Quick check

A cylinder of radius $x$ is inscribed coaxially in a cone of radius $r$ and height $h$ . What is its height?

Quick check

Which is total revenue when $p (x)$ is unit selling price?

Quick check

A derivative equation gives one root outside the feasible domain. What should you do?

Practice Problems

Find two positive numbers with sum 60 maximizing $x y^{3}$ . Then solve the corresponding problem with sum 35 and product $x^{2} y^{5}$ .
Two positive numbers sum to 16. Find the least possible sum of their cubes and justify the global minimum.
Find the best corner cut for an open box made from an 18 cm square, and then from a 45 cm by 24 cm rectangle. State both feasible domains.
Show that a rectangle of greatest area inscribed in a fixed circle is a square.
For a closed cylinder of fixed total surface area, show that greatest volume occurs when height equals diameter. State explicitly that the circular ends are included.
Find the dimensions of a closed cylindrical can of volume 100 cm³ with least total surface area.
A 28 m wire is divided to make a square and a circle. Minimize their combined area, and report the length assigned to each shape.
Show that a cone of greatest volume inscribed in a sphere of radius $R$ has height $4 R /3$ and occupies $8/27$ of the sphere’s volume.
A cone has fixed volume. Minimize its curved surface area and show that its height is $2$ times its radius.
For a cone of fixed slant height, maximize volume and find its semi-vertical angle. Repeat for fixed total surface area, explicitly including the base.
A rectangular tank has depth 2 m and volume 8 m³. Base material costs 70 rupees/m² and side material 45 rupees/m². Find its least construction cost.
A window consists of a rectangle topped by a semicircle, with total outside perimeter 10 m. Find dimensions maximizing its area.
Find the greatest area of an isosceles triangle inscribed in $x^{2} / a^{2} + y^{2} / b^{2} = 1$ , with its vertex at an endpoint of the major axis. Assume $a > b > 0$ .
A cylinder is inscribed in a sphere of radius $R$ . Express volume in terms of cylinder height and find its greatest possible value.
A right circular cylinder is inscribed coaxially in a cone of height $h$ and semi-vertical angle $α$ . Show that its greatest-volume height is $h /3$ .
For a right triangle, a point on the hypotenuse is at perpendicular distances $a, b > 0$ from the two legs. Express the hypotenuse length using its angle and show its minimum is $(a^{2/3} + b^{2/3})^{3/2}$ .

Key Takeaways

• Optimization starts with a correct objective, a constraint, and a feasible domain. • Geometric relationships can reduce a model to one variable. • Minimizing nonnegative squared distance finds the same locations as minimizing distance. • Only feasible stationary points can answer the physical question. • Justify the global comparison and finish with the requested dimensions, values, and units.

Previous · Lesson 4

Testing for Local Extrema

Next · Lesson 6

Absolute Extrema on Closed Intervals

From a Situation to a Function

Numbers Subject to a Fixed Sum

Minimizing a Distance or Its Square

An Area Model Built from Geometry

A Solid Inscribed in Another Solid

A Folded Box and a Feasible Root

Keeping Surface Area and Volume Constraints Distinct

Profit Is Revenue Minus Cost

Check Your Understanding

Quiz

Practice Problems

Key Takeaways