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Use differentiation to maximise area or volume and minimise distance or material. Build a one-variable model, apply physical domains and justify the optimum.
Before you startDifferentiation, stationary points and area/volume formulas
01 / Build the model
A stationary point is a candidate, not an automatic optimum. A negative second derivative proves a local maximum; global claims require the allowed domain too. For a continuous function on a closed bounded interval, compare its stationary points, any non-differentiable interior candidates and both endpoints.
02 / Area with a fixed fence
A rectangular growing plot uses a wall for one side and 36 m of fencing for the other three. Let x be the distance from the wall and y the length parallel to it.
2x + y = 36 ⇒ y = 36 − 2x
A(x) = xy = 36x − 2x²
Physical domain: 0 < x < 18
A′(x) = 36 − 4x is zero at x = 9. Then y = 18 and A = 162 m². Since A″ = −4 throughout, and the area increases before 9 and decreases after it, this is the global maximum on the physical domain.
Move the width yourself. A larger width uses more of the fence twice, leaving less for the long side.
Width x = 9 m and length y = 18 m use 2x + y = 36 m of fence. Area = 162 m². A′(x) = 0 m² per m of width. The maximum is reached at x = 9 m, y = 18 m. The rectangle diagram uses equal scales.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / An open box with square ends
A box has two square end faces of side x metres, a rectangular base and two long side faces. There is no top. Let its length be L metres, and suppose 48 m² of material is available.
Material: 2x² + 3xL = 48
L = (48 − 2x²)/(3x)
Volume: V = x²L = 16x − (2/3)x³
Domain: 0 < x < 2√6
V′(x) = 16 − 2x² gives x = 2√2 after rejecting the negative length. Then L = 8√2/3 and V = 64√2/3 m³. V″(x) = −4x < 0 throughout the physical domain; the volume tends to zero at both limiting degenerate shapes.
04 / Cylinder constraints
A closed cylinder has radius r, height h and surface area 150π cm². Counting two discs and the curved wall gives:
2πr² + 2πrh = 150π
h = 75/r − r
V(r) = 75πr − πr³
0 < r < 5√3
V′(r) = 75π − 3πr² is zero at r = 5 cm. The height is 10 cm and maximum volume is 250π cm³. The derivative changes from positive to negative, and both limiting endpoint volumes are zero.
If instead the volume is fixed at 128π cm³, h = 128/r² and:
S(r) = 2πr² + 256π/r, r > 0
S′(r) = 4πr − 256π/r²
S′(r) = 0 ⇒ r³ = 64 ⇒ r = 4
Then h = 8 cm and S = 96π cm². Since S″(r) = 4π + 512π/r³ > 0 and S grows without bound at both domain ends, this is the global minimum.
05 / A sector with fixed area
A circular sector has area 72 cm², radius r cm and arc length ℓ cm. Its area is rℓ/2: the sector is the same fraction of the circle’s area as its arc is of the circumference.
rℓ/2 = 72 ⇒ ℓ = 144/r
Perimeter P(r) = 2r + 144/r
For a non-degenerate sector smaller than a full circle, r > √(72/π). The stationary condition 2 − 144/r² = 0 gives r = 6√2 cm, which is allowed. P″ = 288/r³ > 0, so this is the global minimum; P = 24√2 cm.
The two straight radii both contribute to the perimeter. The arc length is not the sector’s area.
06 / A rectangle with a semicircular top
A display panel has a rectangular part of width 2r and height h, topped by a semicircle of radius r. Its outside perimeter is 24 cm.
2h + 2r + πr = 24
h = 12 − (1 + π/2)r
A(r) = 2rh + πr²/2
= 24r − (2 + π/2)r²
The domain is 0 < r < 24/(2 + π). Setting A′ = 24 − (4 + π)r to zero gives r = 24/(4 + π), and then h = r. The maximum area is 288/(4 + π) cm². A″ = −(4 + π) < 0, so the quadratic rises then falls across the allowed interval.
07 / Frames with internal divisions
A wire frame contains four columns and three rows of equal small rectangles, each x mm wide and y mm high. Its total width is 4x and height is 3y. There are four full horizontal wires and five full vertical wires.
Total wire: 4(4x) + 5(3y) = 480
y = 32 − (16/15)x
Overall area A = 12xy
= 384x − (64/5)x²
For 0 < x < 30, A′ = 384 − (128/5)x is zero at x = 15 mm. Then y = 16 mm, the frame is 60 by 48 mm and its maximum area is 2880 mm².
Counting only the outer perimeter would give a different constraint and a different answer.
08 / Minimise a squared distance
Find the shortest distance from O to the part of y = 4 − x²/2 with y ≥ 0. A point P has squared distance:
D² = x² + (4 − x²/2)²
= x⁴/4 − 3x² + 16
−2√2 ≤ x ≤ 2√2
Because square root is increasing on non-negative values, minimising D² also minimises D. Differentiate the squared expression:
d(D²)/dx = x³ − 6x = x(x² − 6)
Candidates: x = 0, ±√6
At x = ±√6, y = 1 and D² = 7. At x = 0, D² = 16; at each endpoint D² = 8. Hence the minimum distance is √7, at two points. The identity D² = (x² − 6)²/4 + 7 independently confirms the lower bound.
09 / An overlapping lid
A thin cylindrical container consists of a base disc, a wall of height h, a lid disc and a lid rim 2 cm deep. The common radius is r cm. Ignore thickness and waste. The available material is 78π cm².
2πr² + 2πrh + 4πr = 78π
h = 39/r − r − 2
V(r) = π(39r − r³ − 2r²)
Positive height requires 0 < r < 2√10 − 1. The derivative is π(39 − 3r² − 4r) = −π(r − 3)(3r + 13). Its only positive stationary input is r = 3, giving h = 8 cm and V = 72π cm³.
The derivative is positive before 3 and negative after it. At the maximum, the lid uses πr² + 4πr = 21π cm², which is 21/78 = 7/26 of the material, about 26.9%.
The overlapping rim still needs material. It is not subtracted because it lies over another surface, and it does not increase the container’s internal height in this model.
10 / Extension: a triangular trough
An open trough has two identical right-isosceles triangular ends. Its open top width w is each triangle’s hypotenuse, so each equal side is w/√2 and each triangular area is w²/4. Let the required volume be Q m³ and the length be L m.
Q = (w²/4)L ⇒ L = 4Q/w²
Material S = 2(w²/4) + 2L(w/√2)
= w²/2 + 4√2 Q/w, for w > 0
Write K = 4√2 Q, a positive constant. Then S′ = w − K/w², so the minimum occurs at w³ = K. Since S″ = 1 + 2K/w³ > 0 and S grows without bound at both domain ends, the minimum is global.
w = (4√2 Q)1/3
Smin = (3/2)(4√2 Q)2/3
For example, Q = 8√2 m³ gives w = 4 m, L = 2√2 m and S = 24 m². The open top is excluded from the material count.
11 / Your turn
Justify the optimum and give units wherever dimensions are supplied.
A rectangular plot has 28 m of fence for three sides. Find its maximum area and dimensions.
Let the two short sides each be x, so the other side is 28 − 2x.
A = 28x − 2x², 0 < x < 14
A′ = 28 − 4x = 0 ⇒ x = 7
Other side = 14; maximum area = 98 m².
A″ = −4 and the derivative changes from positive to negative.
Find the maximum and minimum of f(x) = x² + 2x on −2 ≤ x ≤ 1.
Compare the stationary input −1 with both endpoints.
f(−2) = 0, f(−1) = −1, f(1) = 3.
Minimum −1; maximum 3.
A closed cylinder uses 96π cm² of material. Find its maximum volume.
h = 48/r − r and V = 48πr − πr³.
0 < r < 4√3
V′ = 48π − 3πr² = 0 ⇒ r = 4
h = 8; Vmax = 128π cm³.
The derivative changes + to − and the limiting endpoint volumes are zero.
A closed cylinder has volume 54π cm³. Find the least surface area.
h = 54/r² and S = 2πr² + 108π/r.
S′ = 4πr − 108π/r² = 0 ⇒ r³ = 27
r = 3, h = 6, Smin = 54π cm².
S″ = 4π + 216π/r³ > 0 for r > 0, with S → ∞ at both ends.
A sector has area 50 cm². Find its least perimeter, allowing any non-degenerate angle below a full turn.
P = 2r + 100/r, with r > √(50/π).
P′ = 2 − 100/r² = 0 ⇒ r = 5√2
Pmin = 20√2 cm.
The stationary radius is allowed and P″ = 200/r³ > 0.
The rectangle-and-semicircle panel above has perimeter 20 cm instead. Find the maximum area.
A = 20r − (2 + π/2)r².
r = 20/(4 + π), h = r
Amax = 200/(4 + π) cm².
The allowed interval is 0 < r < 20/(2 + π); the stationary point lies inside and A″ < 0.
A frame is divided into two columns and two rows of small rectangles of size x by y cm. Total wire is 120 cm. Find the largest overall area.
Three horizontal wires each have length 2x; three vertical wires each have length 2y.
6x + 6y = 120 ⇒ y = 20 − x
A = 4xy = 80x − 4x²
x = y = 10 ⇒ Amax = 400 cm².
Use 0 < x < 20; the area is a downward quadratic.
Find the shortest distance from O to y = 3 − x², with y ≥ 0.
Minimise D² = x⁴ − 5x² + 9 on −√3 ≤ x ≤ √3.
d(D²)/dx = 2x(2x² − 5)
At x = ±√(5/2), y = 1/2 and D² = 11/4.
Minimum distance = √11/2.
The other stationary value is 9 at x = 0 and the endpoint values are 3. Both closest points are allowed.
An open box has square ends of side x and length L. Its material area is 24 m². Find its maximum volume.
Use 2x² + 3xL = 24.
V = 8x − (2/3)x³, 0 < x < 2√3
V′ = 8 − 2x² = 0 ⇒ x = 2
L = 8/3; Vmax = 32/3 m³.
V″ = −4x < 0 and the limiting endpoint volumes are zero.
A container designer writes 2πr² + 2πrh + 2πr = 32π for a thin cylindrical container and its lid. Explain the three terms, then maximise the container volume and find the fraction of material in the lid.
The last term is a lid rim of depth 1 cm. V = π(16r − r³ − r²).
V′ = π(16 − 3r² − 2r)
= π(2 − r)(3r + 8)
r = 2, h = 5, Vmax = 20π cm³.
Lid area = πr² + 2πr = 8π: fraction 1/4.
The terms count two discs, the main wall and the lid rim. The physical domain is 0 < r < (√65 − 1)/2; the derivative changes + to − at 2.
For S(w) = w²/2 + K/w with K > 0 and w > 0, find the minimum value in terms of K.
The stationary equation is w³ = K.
w = K1/3
Smin = (3/2)K2/3.
S″ = 1 + 2K/w³ > 0 and S tends to infinity at both ends of the domain.
A stationary equation gives x² = 9 when x is a box width. Is reporting x = ±3 a complete contextual answer?
Apply the physical domain after solving the algebra.
No. A positive width requires x = 3; −3 is rejected. You must still verify the other dimensions are positive and justify that the retained candidate gives the required optimum.
12 / Recap
Section 1 of 12 · Build the model