Imagine planning a garden with 40 metres of fencing and wanting the largest rectangular area. A long, thin rectangle wastes space; a square feels roomier; somewhere in between lies the best shape. That is an optimisation problem — a fixed resource and a quantity to make as large (or as small) as possible.
Every optimisation word problem has the same skeleton: the best outcome — maximum area, minimum cost, largest volume, shortest time — under a constraint — limited material, a fixed budget, a given perimeter.
The Plan of Attack
The problem gives you a story, not a graph. Your job is to turn it into a single-variable function and find its peak or valley:
Name the quantity to optimise — call it Q, and write it using variables.
Find the constraint — a relation between those variables (e.g. "perimeter =40").
Reduce to one variable — use the constraint to eliminate the rest.
Differentiate — solve Q′(x)=0 to find the critical points.
Confirm — use Q′′(x)<0 for a maximum or Q′′(x)>0 for a minimum.
Answer the question asked — give the actual dimensions/cost, not just x.
Note
In board exams these problems almost always reduce to a quadratic or cubic. Once Q(x) is written, the calculus is mechanical.
The Garden, Worked
40 m of fencing encloses a rectangle; maximise the area.
Objective:A=lw.
Constraint:2l+2w=40, so l+w=20.
Reduce:w=20−l, giving A(l)=l(20−l)=20l−l2.
Differentiate:A′(l)=20−2l=0⟹l=10.
Confirm:A′′(l)=−2<0, a maximum.
So l=w=10 m — a 10 m × 10 m square.
Watch out
A common slip: solving A′(l)=0 and stopping. Always check max vs min, and answer in the units asked.
The Common Families
| Problem type | Typical objective | Typical constraint |
Eliminating h from the fixed surface area and maximising V=2Sr−πr3 gives S=6πr2, which forces h=2r — the height equals the diameter.
The idea
The surface area is fixed, which links r and h. We use that relation to write the volume as a function of r alone, then maximise with the derivative (standard CBSE method).
Set up
For a right circular cylinder with base radius r and height h:
Method: Fixed-Surface, Maximum-Volume Optimization for Solids of Revolution
This method solves "given a fixed total surface area, find the dimensions that maximize the volume" problems for shapes like cylinders — reducing a two-variable solid-geometry problem to a single-variable calculus problem via the surface-area constraint.
Steps
Step 1: Write both the fixed constraint and the objective in terms of the solid's dimensions
For a right circular cylinder with base radius r and height h, closed at both ends:
S=2πr2+2πrh(fixed),V=πr2h
Step 2: Use the constraint to eliminate one variable
Solve the surface-area equation for h in terms of r and the fixed constant S:
h=2πrS−2πr2
Step 3: Substitute into the volume formula to get a single-variable function
V(r)=πr2⋅2πrS−2πr2=2Sr−πr3
Step 4: Differentiate, solve for the critical radius, and confirm a maximum …
Mistake 1: Using the wrong surface-area formula (forgetting one or both circular ends)
Why it's wrong: a "given surface" cylinder problem like this one assumes a closed can with two circular ends, so S=2πr2+2πrh — using only the lateral surface (S=2πrh) or only one end (S=πr2+2πrh) changes every subsequent step and gives a wrong final ratio between h and r. Correct approach: confirm whether the solid is open or closed at each end before writing the surface-area constraint, and use 2πr2+2πrh for a fully closed cylinder.
Mistake 2: Differentiating V while h is still present, instead of eliminating it first
Why it's wrong: V=πr2h has two independent-looking variables; differentiating with respect to r while treating h as constant ignores that h itself depends on r through the surface constraint, producing an incomplete (and wrong) derivative. Correct approach: always substitute the constraint to write V as a function of r alone before differentiating. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL6 marks
Q.Prove that the area of a right angled triangle of a given hypotenuse is maximum when the triangle is isosceles.
OR
Find the area of the smaller portion enclosed by the curves x2+y2=9 and y2=8x.
›Reveal solutionSolution
Main: with hypotenuse h fixed and one leg x, maximize A(x)=21xh2−x2; the maximum occurs at x=h/2, giving equal legs. OR: find the intersection points of the circle and parabola, then integrate.
Main question. Let the hypotenuse have fixed length h, and let one leg be x (0<x<h); the other leg is h2−x2.
Area: A(x)=21xh2−x2. It's easier to maximize A2 (same maximizer since A>0):
A2=41x2(h2−x2)=41(h2x2−x4).
dxd(A2)=41(2h2x−4x3)=2x(h2−2x2).
Setting this to 0 (for x>0): h2−2x2=0⟹x2=2h2⟹x=2h.
Check it's a maximum: dx2d2(A2)x=h/2=2h2−6x2x=h/2=2h2−3h2<0, confirming a maximum (also A→0 at both endpoints x=0,h, consistent with an interior max).
At x=h/2, the other leg is h2−h2/2=h2/2=h/2 — equal to the first leg. So the area is maximum exactly when the two legs are equal, i.e. the triangle is isosceles.
OR question. Find the intersection of x2+y2=9 (circle, radius 3) and y2=8x (parabola).
Substitute y2=8x into the circle: x2+8x=9⟹x2+8x−9=0⟹(x+9)(x−1)=0. Since x≥0 (parabola), x=1, giving y2=8⟹y=±22.
The smaller enclosed region is symmetric about the x-axis, bounded by the parabola from x=0 to 1 and by the circle from x=1 to 3. Its area is
Area=2[∫018xdx+∫139−x2dx].
First integral:∫018xdx=8[32x3/2]01=328=342.