9E - Maximum and minimum problems

9E - Maximum and minimum problems

A fixed amount of material can make a tall narrow container or a short wide one. You will turn a practical constraint into a function of one variable, use calculus to find its best value and check that the answer is possible in the original problem.

Building a one-variable model

If a closed cylindrical can has a fixed surface area, does making its radius larger always increase its capacity?

A larger radius also uses more material in the circular ends, leaving less for the height. Both dimensions affect the volume, so we need to connect them before differentiating.

Objective and constraint

The objective is the quantity being maximised or minimised. A constraint is a relation restricting the possible dimensions. Use the constraint to express the objective in one variable, and state the feasible domain: the values that make sense in the problem.

For a cylinder of radius rr and height hh, the volume is V=πr2hV=\pi r^2h. Its curved surface unwraps into a rectangle of width 2πr2\pi r and height hh, so that area is 2πrh2\pi rh. Each circular face adds πr2\pi r^2.

[DIAGRAM: asset_name: Surface constraints for closed and open cylinders; asset_slug: cylinder-models; caption: Removing a lid changes the surface constraint while the volume formula stays the same; description: Two labelled cylinders show the radius r and height h. The closed cylinder has two circular faces and total surface area 2 pi r squared plus 2 pi r h. The open cylinder has one circular face and surface area pi r squared plus 2 pi r h. Both have volume pi r squared h and require positive radius and height.]
Removing a lid changes the surface constraint while the volume formula stays the same

For a cone, the curved surface area πrℓ\pi r\ell, where ℓ\ell is the slant height, is supplied on the formulae sheet. For any solid, decide which faces are present before writing its constraint.

Worked example 1: a closed cylinder

Question. A closed cylinder has total surface area 96π cm296\pi\,\mathrm{cm}^2. Express its volume in terms of rr and state the feasible domain.

Plan. Count both circular ends, express hh in terms of rr, then substitute into the volume.

2πr2+2πrh=96πsurface constraintr2+rh=48divide by 2πh=48r−reliminate the heightV=πr2(48r−r)=48πr−πr3substitute into volume\begin{aligned} 2\pi r^2+2\pi rh&=96\pi &&\text{surface constraint}\\ r^2+rh&=48 &&\text{divide by }2\pi\\ h&=\frac{48}{r}-r &&\text{eliminate the height}\\ V&=\pi r^2\left(\frac{48}{r}-r\right)=48\pi r-\pi r^3 &&\text{substitute into volume} \end{aligned}

A real container requires r>0r>0 and h>0h>0. Thus 48−r2>048-r^2>0, giving 0<r<430<r<4\sqrt3.

Answer. V=48πr−πr3V=48\pi r-\pi r^3, with 0<r<430<r<4\sqrt3.

Check. At r=2r=2, the model gives h=22h=22. The surface area is 8π+88π=96π8\pi+88\pi=96\pi, and both the direct volume π(2)2(22)\pi(2)^2(22) and the model give 88π cm388\pi\,\mathrm{cm}^3.

Keep the same amount of material but remove the lid. Only the number of circular faces changes; do not retain the closed-cylinder constraint.

Worked example 2: an open cylinder

Question. An open cylindrical container has one circular base and surface area 96π cm296\pi\,\mathrm{cm}^2. Express its volume in terms of rr and state the feasible domain.

Plan. Replace two circular faces by one, then eliminate hh as before.

πr2+2πrh=96πh=96−r22r=48r−r2V=πr2(48r−r2)=48πr−πr32\begin{aligned} \pi r^2+2\pi rh&=96\pi\\ h&=\frac{96-r^2}{2r}=\frac{48}{r}-\frac r2\\ V&=\pi r^2\left(\frac{48}{r}-\frac r2\right) =48\pi r-\frac{\pi r^3}{2} \end{aligned}

The conditions r>0r>0 and h>0h>0 give r2<96r^2<96.

Answer. V=48πr−12πr3V=48\pi r-\frac12\pi r^3, with 0<r<460<r<4\sqrt6.

Check. At r=2r=2, h=23h=23, giving surface area 4π+92π=96π4\pi+92\pi=96\pi. The direct volume is 92π92\pi, matching 96π−4π96\pi-4\pi from the model.

The domain comes from the dimensions, rather than from the algebraic expression for volume alone.

Finding and justifying an optimum

A stationary value is a candidate for an optimum, but the feasible domain decides which candidates can be used.

Differentiate the objective with respect to its remaining variable. Solve for zero derivative, discard infeasible solutions, and justify the nature of each remaining point. On a closed interval, compare all relevant stationary values and endpoint values. On an open interval, a limiting value at an excluded endpoint is not an attained optimum.

Worked example 3: a feasible stationary maximum

Question. Find the radius giving the greatest value of V=48πr−πr3V=48\pi r-\pi r^3 for 0<r<430<r<4\sqrt3, and justify your answer.

Plan. Find the positive stationary radius, then check the gradient over the feasible domain.

dVdr=48π−3πr2=3π(16−r2)=0⟹r=±4.\frac{dV}{dr}=48\pi-3\pi r^2=3\pi(16-r^2)=0 \quad\Longrightarrow\quad r=\pm4.

Only r=4r=4 is feasible. The second derivative is

d2Vdr2=−6πr,d2Vdr2∣r=4=−24π<0,\frac{d^2V}{dr^2}=-6\pi r,\qquad \left.\frac{d^2V}{dr^2}\right|_{r=4}=-24\pi<0,

so this is a local maximum. On the full feasible interval, dV/dr>0dV/dr>0 before 44 and dV/dr<0dV/dr<0 after 44, establishing the greatest volume there.

Answer. r=4 cmr=4\,\mathrm{cm} gives the greatest volume.

Check. The volume approaches zero as rr approaches either feasible-domain boundary, while V(4)=128π>0V(4)=128\pi>0. The boundaries themselves do not describe containers with positive radius and height.

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