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Applications of the Derivative · Topic 18 of 23

Applied Optimization

Applied optimization is one recipe: use the constraint to write the quantity as a single function f(x)f(x) on an explicit domain, then find its absolute maximum or minimum. Solving f′(x)=0f'(x)=0 only produces candidates — the marks live in the domain and in the justification that a candidate is global.

5 min readCore — on every final7 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Reduce to one variable first

    Write the objective, write the constraint as an equation, solve it for one variable and substitute. Differentiating A=xyA=xy with two live variables is the fastest way to lose the whole question.

  2. 2

    The domain carries marks

    Every length built from xx must be non-negative, not just xx itself. Cutting corner squares of side xx from a 30×1630\times16 sheet gives 0≤x≤80\le x\le 8, not x>0x>0.

  3. 3

    Keep the degenerate endpoints

    A box of height 00 is not a box, but admitting x=0x=0 and x=8x=8 costs nothing and turns an open interval into a closed bounded one — the Extreme Value Theorem for free.

  4. 4

    A critical number is only a candidate

    f′(c)=0f'(c)=0 never by itself says maximum. Justify: Closed Interval Method on [a,b][a,b], the full sign of f′f' across II, or f′′>0f''>0 on all of II — not just f′′(c)>0f''(c)>0.

  5. 5

    Minimise the square of a distance

    To find the nearest point on a curve, minimise d2d^2 rather than dd. Since  \sqrt{\ } is increasing, the minimiser is identical, and you avoid a chain rule through a radical.

51015202530100200300400xymax A = 312.5 at x = 12.5x = 0x = 25A(x) = x(50 − 2x)feasible: 0 ≤ x ≤ 25
Fifty metres of fence on three sides of a rectangle: A(x)=x(50−2x)A(x)=x(50-2x). The shaded strip is the feasible domain 0≤x≤250\le x\le 25 — outside it the formula still returns a number, but there is no field. The maximum 312.5 m2312.5\ \text{m}^2 sits at x=12.5x=12.5, y=25y=25, and both endpoints give A=0A=0.
20406080100200400600800xyall circle: 795.8 = the maximumA′ = 0: the minimum, 350.1all square: 625A(x), square + circle
A 100100 cm wire cut into a square (using xx cm) and a circle. A′′=18+12π>0A''=\frac18+\frac{1}{2\pi}>0 everywhere, so the interior critical number x=400π+4≈56x=\frac{400}{\pi+4}\approx56 is the minimum area; the maximum is forced to the endpoint x=0x=0. Report x≈56x\approx56 as "the optimum" and part (b) is simply wrong.
02

Formulas

What to have memorised

  • Closed Interval Method

    max⁡[a,b]f=max⁡{f(a), f(b), f(c1),…,f(cn)}\max_{[a,b]}f=\max\bigl\{f(a),\,f(b),\,f(c_1),\dots,f(c_n)\bigr\}

    Needs ff continuous on a closed bounded [a,b][a,b]; the cic_i are the critical numbers inside.

  • Global concavity test

    f′(c)=0  and  f′′>0  on all of I ⟹ f(c)=min⁡Iff'(c)=0\ \text{ and }\ f''>0\ \text{ on all of }I\ \Longrightarrow\ f(c)=\min_I f

    Flip both signs for an absolute maximum. f′′(c)>0f''(c)>0 alone gives only a local minimum.

  • Closest point of y=f(x)y=f(x) to (p,q)(p,q)

    g(x)=(x−p)2+(f(x)−q)2g(x)=(x-p)^2+\bigl(f(x)-q\bigr)^2

    Minimise g=d2g=d^2; take the square root only if the distance itself is asked for.

  • Enclosure with a river on one side

    A=x(L−2x),0≤x≤L2A=x(L-2x),\qquad 0\le x\le\tfrac{L}{2}

    From 2x+y=L2x+y=L. The best field has x=L4x=\tfrac{L}{4}, twice as wide as it is deep.

  • Open box from an a×ba\times b sheet

    V=x(a−2x)(b−2x),0≤x≤12min⁡(a,b)V=x(a-2x)(b-2x),\qquad 0\le x\le\tfrac12\min(a,b)

    xx is both the corner square cut out and the height of the box.

  • Cheapest can of fixed volume

    C=cs(2πrh)+ct(2πr2),  πr2h=V ⟹ h=2ctcs rC=c_s(2\pi rh)+c_t(2\pi r^2),\ \ \pi r^2h=V\ \Longrightarrow\ h=\frac{2c_t}{c_s}\,r

    Equal side and lid costs give the familiar h=2rh=2r.

  • Least time across an interface

    T(x)=a2+x2v1+b2+(d−x)2v2T(x)=\frac{\sqrt{a^2+x^2}}{v_1}+\frac{\sqrt{b^2+(d-x)^2}}{v_2}

    T′(x)=0T'(x)=0 is Snell's law sin⁡θ1v1=sin⁡θ2v2\frac{\sin\theta_1}{v_1}=\frac{\sin\theta_2}{v_2}.

03

Solve an applied optimization problem

The steps, in order
  1. 1

    Draw and label the picture: a symbol for every length, radius and angle, with units recorded.

  2. 2

    Write the objective QQ as a formula, and the constraint as an equation relating the variables.

  3. 3

    Solve the constraint for one variable, substitute, and reduce to Q=f(x)Q=f(x).

  4. 4

    State the domain: every length built from xx must be non-negative. Keep degenerate endpoints if ff reaches them.

  5. 5

    Find the critical numbers in the domain — where f′(x)=0f'(x)=0 and where f′f' fails to exist.

  6. 6

    Justify globally: Closed Interval Method on [a,b][a,b], or the sign of f′f' or f′′f'' across the whole open domain.

  7. 7

    Convert back to what was asked — dimensions, cost, area, time — and attach units.

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    Differentiating A=xyA=xy as it stands.

    ✓ Right

    Use the constraint first: A(x)=x(400−2x)A(x)=x(400-2x), then differentiate.

    Why: Two variables and one equation is not yet a calculus problem.

  • ✗ Wrong

    "f′(c)=0f'(c)=0, so this is the maximum."

    ✓ Right

    cc is a candidate. Name a justification: Closed Interval Method, sign of f′f', or global concavity.

    Why: A stationary point can be a minimum, or beaten by an endpoint.

  • ✗ Wrong

    f′′(c)>0f''(c)>0, so f(c)f(c) is the global minimum.

    ✓ Right

    That gives a local minimum. Upgrade it with f′′>0f''>0 on the whole interval.

    Why: The Second Derivative Test checks one point and concludes about one point.

  • ✗ Wrong

    Solving for the interior critical number and stopping.

    ✓ Right

    Evaluate the endpoints too. For a wire cut into a square and a circle, the maximum area is at an endpoint.

  • ✗ Wrong

    Writing the domain as "x>0x>0" out of habit.

    ✓ Right

    Intersect every requirement: for a 30×1630\times16 sheet the domain is [0,8][0,8].

    Why: Derived lengths must be non-negative too.

  • ✗ Wrong

    Answering "x=103x=\tfrac{10}{3}" and stopping.

    ✓ Right

    The question asked for a volume, a cost or the dimensions. Convert back and attach units.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    A farmer has 400400 m of fencing for a rectangular field beside a straight river; no fence is needed along the river. Find the dimensions of largest area and justify that it is the absolute maximum.

    Hint

    Let xx be each of the two sides perpendicular to the river, and use all 400400 m of fence.

    Show answer

    Answer

    100100 m deep by 200200 m along the river; maximum area 20 000 m220\,000\ \text{m}^2.

    Steps

    With yy parallel to the river, 2x+y=4002x+y=400, so y=400−2xy=400-2x and

    A(x)=x(400−2x),0≤x≤200.A(x)=x(400-2x),\qquad 0\le x\le 200.

    AA is a polynomial on a closed bounded interval, so the Closed Interval Method applies. A′(x)=400−4x=0A'(x)=400-4x=0 gives the only critical number x=100x=100.

    A(0)=0A(0)=0, A(100)=100⋅200=20 000A(100)=100\cdot200=20\,000, A(200)=0A(200)=0. The interior candidate wins, so x=100x=100 m and y=200y=200 m.

  2. Q2easy

    An open-topped box is made from a 1212 cm square of cardboard by cutting a square of side xx from each corner and folding up the flaps. Find the xx that maximises the volume, and state that volume.

    Show answer

    Answer

    x=2x=2 cm; maximum volume 128 cm3128\ \text{cm}^3, from an 8×8×28\times8\times2 cm box.

    Steps

    The base is (12−2x)(12-2x) by (12−2x)(12-2x) and the height is xx, so

    V(x)=x(12−2x)2,0≤x≤6.V(x)=x(12-2x)^2,\qquad 0\le x\le 6.

    V′(x)=(12−2x)2−4x(12−2x)=(12−2x)(12−6x)V'(x)=(12-2x)^2-4x(12-2x)=(12-2x)(12-6x)

    vanishes at x=6x=6 and x=2x=2; only x=2x=2 is interior. Closed Interval Method: V(0)=0V(0)=0, V(2)=2⋅82=128V(2)=2\cdot8^2=128, V(6)=0V(6)=0.

  3. Q3medium

    A box with an open top and a square base must hold 32 000 cm332\,000\ \text{cm}^3. Find the dimensions that use the least material, and justify that your answer is the global minimum.

    Hint

    Eliminate the height with the volume constraint. The domain is open, so the Closed Interval Method is not available.

    Show answer

    Answer

    Base 40 cm×40 cm40\ \text{cm}\times40\ \text{cm}, height 2020 cm; least surface area 4800 cm24800\ \text{cm}^2.

    Steps

    Base side xx, height hh, with x2h=32 000x^2h=32\,000, so h=32 000/x2h=32\,000/x^2 and

    S(x)=x2+4xh=x2+128 000x,x>0.S(x)=x^2+4xh=x^2+\frac{128\,000}{x},\qquad x>0.

    S′(x)=2x−128 000x2=0S'(x)=2x-\dfrac{128\,000}{x^2}=0 gives x3=64 000x^3=64\,000, so x=40x=40 is the only critical number.

    S′′(x)=2+256 000x3>0S''(x)=2+\dfrac{256\,000}{x^3}>0 for every x>0x>0, so SS is concave up on the whole domain and x=40x=40 is the absolute minimum. Then h=32 000/1600=20h=32\,000/1600=20 and S=1600+3200=4800 cm2S=1600+3200=4800\ \text{cm}^2.

  4. Q4medium

    Find the point or points on the parabola y=x2y=x^2 closest to (0,3)(0,3), and state the shortest distance.

    Hint

    Minimise the square of the distance, not the distance.

    Show answer

    Answer

    Two points, (±102, 52)\left(\pm\frac{\sqrt{10}}{2},\ \frac52\right); shortest distance 112≈1.66\frac{\sqrt{11}}{2}\approx1.66.

    Steps

    Since  \sqrt{\ } is increasing, dd and g=d2g=d^2 are minimised at the same xx:

    g(x)=x2+(x2−3)2=x4−5x2+9.g(x)=x^2+(x^2-3)^2=x^4-5x^2+9.

    g′(x)=4x3−10x=2x(2x2−5)=0g'(x)=4x^3-10x=2x(2x^2-5)=0 at x=0x=0 and x=±102x=\pm\frac{\sqrt{10}}{2}.

    Sign of g′g' across the line: −,+,−,+-,+,-,+, so gg has a local maximum at x=0x=0 and ties for its absolute minimum at x=±102x=\pm\frac{\sqrt{10}}{2}. There x2=52x^2=\frac52, so g=254−252+9=114g=\frac{25}{4}-\frac{25}{2}+9=\frac{11}{4} and d=112d=\frac{\sqrt{11}}{2}.

  5. Q5medium

    A Norman window is a rectangle topped by a semicircle whose diameter is the rectangle's upper edge. The perimeter of the whole window is 1010 m. Find the radius rr and the rectangle's height hh that maximise the area, and state that area.

    Hint

    Only the semicircular arc is on the boundary — the top edge of the rectangle is interior.

    Show answer

    Answer

    r=h=104+π≈1.40r=h=\dfrac{10}{4+\pi}\approx1.40 m; maximum area 504+π≈7.00 m2\dfrac{50}{4+\pi}\approx7.00\ \text{m}^2.

    Steps

    The rectangle is 2r2r wide, so the perimeter is 2r+2h+πr=102r+2h+\pi r=10, giving h=10−(2+π)r2h=\frac{10-(2+\pi)r}{2}. Then

    A(r)=2rh+12πr2=10r−4+π2r2,0≤r≤102+π.A(r)=2rh+\tfrac12\pi r^2=10r-\frac{4+\pi}{2}r^2,\qquad 0\le r\le\frac{10}{2+\pi}.

    A′(r)=10−(4+π)r=0A'(r)=10-(4+\pi)r=0 at r=104+π≈1.40r=\frac{10}{4+\pi}\approx1.40, which is interior. A′′=−(4+π)<0A''=-(4+\pi)<0 everywhere, so this is the absolute maximum.

    Then h=104+π=rh=\frac{10}{4+\pi}=r and A=1004+π−504+π=504+πA=\frac{100}{4+\pi}-\frac{50}{4+\pi}=\frac{50}{4+\pi}.

  6. Q6hard

    A rower sits 66 km offshore from the nearest shore point AA. Her destination BB lies 2020 km along the straight shore from AA. She rows at 33 km/h and walks at 55 km/h. Where should she land to reach BB in the least time, and how long does the trip take?

    Hint

    Let xx be the distance from AA to the landing point; each leg contributes distance divided by speed.

    Show answer

    Answer

    Land 4.54.5 km from AA (then walk 15.515.5 km); minimum time 5.65.6 hours.

    Steps

    T(x)=36+x23+20−x5,0≤x≤20.T(x)=\frac{\sqrt{36+x^2}}{3}+\frac{20-x}{5},\qquad 0\le x\le 20.

    T′(x)=x336+x2−15=0T'(x)=\dfrac{x}{3\sqrt{36+x^2}}-\dfrac15=0 gives 5x=336+x25x=3\sqrt{36+x^2}, so 25x2=324+9x225x^2=324+9x^2 and x=4.5x=4.5. Squaring can invent roots, so check the unsquared equation: 56.25=7.5\sqrt{56.25}=7.5 and 4.53(7.5)=15\frac{4.5}{3(7.5)}=\frac15. ✓

    Closed Interval Method: T(0)=2+4=6T(0)=2+4=6, T(4.5)=2.5+3.1=5.6T(4.5)=2.5+3.1=5.6, T(20)=4363≈6.96T(20)=\frac{\sqrt{436}}{3}\approx6.96.

06

On the exam

How this topic is marked

  • Most of the marks are in the set-up: the constraint used to eliminate a variable, the domain written down, and one sentence of justification. The differentiation is usually a single line.

  • Name the justification you are using — "Closed Interval Method", "A′′<0A''<0 on the whole interval", "the only critical number of a continuous function on II". "f′(c)=0f'(c)=0, so it is the maximum" earns nothing.

  • Finish by answering the question that was asked — dimensions, cost, area or time — with units, and check the constraint is satisfied.

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