Connected rates and optimisation
Connect changing dimensions and net rates, then justify constrained optima using the original model and feasible domain.
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A relation connects rates through the chain rule
A rate needs both a changing quantity and the variable it changes with. For a sphere,
If V and r change with time t, differentiation gives
The geometric derivative dV/dr = 4π r² is not the volume’s time rate. Its units are cm² when r is in centimetres; multiplying by dr/dt in cm/s gives dV/dt in cm³/s.
At r = 3 cm with dV/dt = 12π cm³/s,
Substitute the instantaneous radius after differentiating the relationship. Replacing r by the constant 3 first would erase the changing radius. If the volume were decreasing at 12π cm³/s, its signed rate would be -12π and the radius rate would be -1/3 cm/s; the positive speed of decrease would still be 1/3 cm/s.
Check your understanding 1 Keep the units of both rates
A square has side s cm and area A cm². Its side grows at 0.2 cm/s. Find dA/dt when s = 5 cm, and explain why giving dA/ds alone would answer a different question.
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A = s², so dA/dt = 2s(ds/dt). At the stated instant this is 2(5)(0.2) = 2 cm²/s. The geometric derivative dA/ds = 2s = 10 cm describes area change per side-length change; it still needs multiplication by the side’s time rate.
Several dimensions may change at the same instant
For a rectangle with changing length l and width w,
Both product-rule contributions matter. At one instant, suppose l = 6 m, w = 4 m, dl/dt = 0.3 m/s and dw/dt = -0.1 m/s. Then
The area is increasing at that instant even though the width is decreasing. A dimension’s current value is not evidence that it stays constant; set a rate to zero only when the model says that dimension is fixed.
For an inflow and an outflow, first find the net volume rate: inflow minus outflow. An incoming rate alone does not equal the stored volume’s rate of change when liquid is also leaving.
Optimisation starts with a constrained model
Optimisation asks for the greatest or least allowed value. State the variable and feasible domain, use the constraint to express the objective in one variable, then justify the chosen optimum. A derivative calculation is part of that argument.
A rectangle has perimeter 40 m. Let its width be x metres and length be l metres. The constraint 2x + 2l = 40 gives l = 20-x. Positive dimensions require 0 < x < 20. Its area is
Differentiate: A'(x) = 20-2x = 0 at x = 10. Since A'' = -2 < 0, this is a local maximum. The form
also proves it is the global maximum, attained at x = 10 within the domain. The dimensions are 10 m by 10 m and the area is 100 m². They satisfy the original perimeter.
The constraint was essential: treating width and length as independent would not give the one-variable area function for these allowed rectangles. Completing the square supplies a different verification of the calculus result.
Feasible boundaries can replace a stationary optimum
If the width must instead satisfy 3 ≤ x ≤ 8, the stationary value 10 is unavailable. The derivative is positive throughout the allowed interval, so the greatest area is at x = 8, with length 12 m and area 96 m².
A stationary point outside the feasible domain cannot be selected just because its derivative is zero. Boundary inclusion also matters: if the restriction were 3 ≤ x < 8, the areas would approach 96 m² without attaining it, so there would be no greatest area under that restriction.
Check your understanding 2 The constraint selects the usable candidate
Keep the rectangle’s 40 m perimeter, but require 12 ≤ x ≤ 16. Find its greatest area and dimensions. What changes if the lower bound becomes 12 < x ≤ 16?
Show answer
A'(x) = 20-2x < 0 throughout [12,16], so area decreases as x increases. The greatest area is at x = 12, with length 8 m and area 96 m². The stationary value 10 lies outside this domain.
If x = 12 is excluded, the area approaches 96 m² as x approaches 12 from above but never attains it. There is then no greatest area, even though there is a finite upper bound.
Try it yourself 1
Track the liquid level
Problem
A fixed cone-shaped tank has its point at the bottom, total height 12 cm and top radius 6 cm. Liquid enters at 9π cm³/s and leaves at 3π cm³/s. At an instant when its depth h is 6 cm, find the rates at which the depth and liquid-surface radius r increase. Assume a horizontal liquid surface and no other volume changes. Use V = π r²h/3. Explain why treating the instantaneous radius as a constant before differentiating gives a wrong height rate.
Write your answer, then compare it with the solution
Hints
Hint 1
Decide which dimensions belong to the changing liquid
Compare the small and full cross-sections, then find the net rate of stored volume. The tank’s fixed top radius is a different quantity from the liquid’s surface radius.
Show full solution
Use the tank's shape
Method
Relate the liquid radius to its depth before differentiating.Reason
The straight sides give similar cross-sections, so the two changing dimensions are linked.Working
Answer
r/h = 6/12 = 1/2, so r = h/2 and V = π h³/12. The net volume rate is 9π-3π = 6π cm³/s.Connect volume, height and radius rates
Method
Differentiate the reduced volume function with respect to time.Reason
The depth is changing; its instantaneous value is substituted after the time derivative is formed.Working
Answer
dV/dt = (π h²/4)(dh/dt). At h = 6, 6π = 9π(dh/dt), so dh/dt = 2/3 cm/s. Since r = h/2, dr/dt = 1/3 cm/s.Check the alternative differentiation
Method
Keep both dimensions changing in the original volume relation.Reason
Holding the instantaneous radius fixed would omit a real contribution to the volume change.Working
Answer
dV/dt = (π/3)[2rh(dr/dt) + r²(dh/dt)]. With r = 3, h = 6 and dr/dt = (dh/dt)/2, this again gives dV/dt = 9π(dh/dt). Freezing r = 3 before differentiation would instead give 3π(dh/dt) and the incorrect height rate 2 cm/s.
Quick check
The net rate is a volume rate in cm³/s; both requested answers are length rates in cm/s. The liquid radius changes with its height, while the tank’s full dimensions stay fixed. Compare the net flow, geometric constraint and chain-rule factors separately.
Try it yourself 2
Choose the container dimensions
Problem
An ideal cylindrical container has a base and curved side, with no lid. Its volume is 16π cm³. Neglect material thickness, joins and waste. Using radius r cm and height h cm, find the dimensions that minimise material area. Then solve the problem again if the radius must be at most 2 cm. Use volume π r²h, base area π r² and curved-side area 2π rh. Justify a global minimum and check the original volume in the restricted design.
Write your answer, then compare it with the solution
Hints
Hint 1
Separate the fixed volume from the area to minimise
Use the volume constraint to remove one dimension from the material-area expression. The open top changes which surfaces count, and the radius restriction changes which designs are allowed.
Show full solution
Form a one-variable objective
Method
Use the fixed volume to express height in terms of radius.Reason
Every feasible design must preserve the stated volume while its area changes.Working
Answer
h = 16/r² for r > 0. The material area is A = π r² + 2π rh = π r² + 32π/r. There is one circular base, not two, because the container has no lid.Find and justify the unrestricted minimum
Method
Differentiate the area and inspect its sign on the positive-radius domain.Reason
A zero derivative supplies a candidate; decreasing before it and increasing afterwards proves the global minimum.Working
Answer
A' = 2π r-32π/r² = (2π/r²)(r³-16). It is negative for 0 < r < cube root of 16 and positive above that value. Thus the minimum has r = cube root of 16 = 2 cube root of 2 cm and h = r. The minimum area is 3π r² = 12π cube root of 4 cm².Apply the radius restriction
Method
Reconsider the objective on 0 < r ≤ 2.Reason
The original stationary radius exceeds the permitted upper bound.Working
Answer
The derivative is negative throughout this restricted domain, so the minimum is at r = 2 cm. Then h = 4 cm and A = 20π cm². The volume check is π(2)²(4) = 16π cm³, as required.
Quick check
The unconstrained and radius-limited problems have different feasible minima. Both use the same fixed-volume relation and the correct open-container area. Compare the surfaces counted, one-variable domain, derivative sign and final dimensions rather than accepting a stationary value without its model.
For connected rates, identify every changing quantity and differentiate the relationship before substituting an instant’s values. For optimisation, identify the objective and feasible domain before selecting candidates. Finish with the original setting, appropriate units and a reason the selected value answers the actual question.
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