PYQ Vault

JEE Mains Maths · Application of Derivatives

Optimisation Problems

Word problems solved by the derivative: the largest area or volume under a constraint, and the shortest distance between a point, a line and a curve.

Why this matters

Eighteen PYQs, fourteen of them multiple choice. Fourteen maximise or minimise an area or a volume — a box folded from a sheet, a wire cut into two shapes, a rectangle inside a curved region; four find a shortest distance or perimeter. Two ideas cover the page.

Concept 1 of 2: Largest area or volume

Name one variable, write the quantity to optimise in terms of it using the constraint, and note the variable's allowed range. Then set the derivative to zero and check that the point is a maximum (or minimum) inside that range. For a wire cut into two shapes, let one piece be the variable and the other the remainder.

Definition

  • One variable; use the constraint to remove the others.
  • Solve Q′(x)=0Q'(x)=0 inside the allowed range; check with Q′′Q'' or the endpoints.
  • Square of perimeter pp: area p216\frac{p^2}{16}; circle of circumference pp: area p24π\frac{p^2}{4\pi}.
  • Open box from an a×ba\times b sheet: V=x(a−2x)(b−2x)V=x(a-2x)(b-2x).

The method

Q=Q(x),Q′(x)=0,Q′′(x)<0 ⇒ maximumQ=Q(x),\quad Q'(x)=0,\quad Q''(x)<0\ \Rightarrow\ \text{maximum}

Worked example

Find the largest area of a rectangle with perimeter 20.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q64Moderate

Example 1 · Application of Derivatives · Optimisation Problems

A square piece of tin of side 30 cm30\text{ }cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm2cm^{2} ) is equal to

Stay inside the allowed range

The cut in a box must be less than half the sheet's side, and a wire piece cannot be negative. A root of Q′Q' outside that range is not the answer.

Concept 2 of 2: Shortest distances

The shortest distance from a curve to a line it does not meet is along a normal, so the nearest point is where the tangent is parallel to the line. From a point to a curve, minimise the squared distance, which avoids the square root. From a circle, work from its centre and subtract the radius. For the least sum of two distances to a point on a line, reflect one point in the line.

Definition

  • Curve to line (not meeting): tangent parallel to the line.
  • Point to curve: minimise D2D^2.
  • Circle: distance from the centre, minus the radius.
  • Least PA+PBPA+PB with PP on a line: reflect AA and join to BB.

Nearest point to a line

f′(x0)=slope of the linef'(x_0)=\text{slope of the line}

Worked example

Find the shortest distance from (0,3)(0,3) to y=x2y=x^2.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 2 · Q66Moderate

Example 2 · Application of Derivatives · Optimisation Problems

If PP is a point on the parabola y=x2+4y=x^{2}+ 4 which is closest to the straight line y=4x−1y= 4x- 1, then the co-ordinates of PP are :

Only if they do not meet

The parallel-tangent method assumes the curve and the line do not cross; if they meet, the shortest distance is 0. Check with a discriminant first.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Largest area or volume

    The method

    Q=Q(x),Q′(x)=0,Q′′(x)<0 ⇒ maximumQ=Q(x),\quad Q'(x)=0,\quad Q''(x)<0\ \Rightarrow\ \text{maximum}
  • Shortest distances

    Nearest point to a line

    f′(x0)=slope of the linef'(x_0)=\text{slope of the line}

Watch out for (2)

Test yourself on Application of Derivatives

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.