Maximum and Minimum Problems

The sign of \dfrac{d^2V}{dx^2} tells you whether it is a maximum or a minimum.

1. \dfrac{d^2V}{dx^2}=-6 is negative, so the stationary point is a maximum
2. 4 is the value of x, and -6 is the value of \dfrac{d^2V}{dx^2}, not of V
3. The value of V there is 58
4. Answer: The maximum value of V is 58

Differentiate C and set \dfrac{dC}{dx}=0.

1. \dfrac{dC}{dx}=x-30
2. At a minimum, x-30=0
3. \dfrac{d^2C}{dx^2}=1>0, so this is a minimum
4. Answer: x=30

Multiply out the brackets to write the area as a quadratic before you differentiate.

1. A=(x+2)(16-x)=16x-x^2+32-2x
2. A=-x^2+14x+32
3. \dfrac{dA}{dx}=-2x+14=0, so x=7
4. A=(7+2)(16-7)=9\times9=81
5. Answer: 81 cm^2

Count how many lengths of x and y the fencing makes, use this to write y in terms of x, then substitute into A=xy.

1. The fencing is three widths and two lengths: 3x+2y=120
2. y=\dfrac{120-3x}{2}=60-1.5x
3. A=xy=x(60-1.5x)
4. Answer: A=60x-\frac32x^2

A cuboid has four edges of each length, so add up 4\times3x, 4\times x and 4\times h.

1. Total edge length: 4(3x)+4x+4h=80
2. 16x+4h=80, so h=20-4x
3. V=3x\times x\times h=3x^2(20-4x)
4. Answer: V=60x^2-12x^3

Sketch the curve through its maximum and minimum points, then think about where a horizontal line y=k crosses it.

1. The curve rises to the maximum (-1,\ 7), falls to the minimum (3,\ -25), then rises for ever
2. y=10 is above the maximum, so it meets the curve only once (to the right of the minimum)
3. y=0 lies between -25 and 7, so it meets the curve three times
4. 7 is only a local maximum: for large x the curve goes above 7
5. y=-30 is below the minimum, so it meets the curve only once
6. Answer: \mathrm{f}(x)=10 has exactly one solution

Write \dfrac{25}{x} as 25x^{-1} and differentiate twice, taking care with the signs.

1. y=20-4x-25x^{-1}
2. \dfrac{dy}{dx}=-4+25x^{-2}
3. \dfrac{d^2y}{dx^2}=-50x^{-3}=-\dfrac{50}{x^3}
4. When x=2.5: -\dfrac{50}{15.625}=-3.2
5. Negative second derivative means a maximum
6. Answer: \dfrac{d^2y}{dx^2}=-\dfrac{50}{x^3}=-3.2, which is negative, so it is a maximum

Differentiate, set \dfrac{dP}{dx}=0 to find x, then substitute that x back into P.

1. \dfrac{dP}{dx}=96-6x^2
2. At a maximum, 96-6x^2=0
3. x^2=16, so x=4 (the number of consoles cannot be negative)
4. P=96\times4-2\times4^3=384-128=256
5. The question asks for the maximum value of P, not x
6. Answer: 256 (a profit of £256 thousand)

Write \dfrac{7200}{x} as 7200x^{-1}, differentiate, set \dfrac{dC}{dx}=0, then substitute back into C.

1. C=2x+7200x^{-1}
2. \dfrac{dC}{dx}=2-7200x^{-2}
3. 2-\dfrac{7200}{x^2}=0, so x^2=3600 and x=60 (a speed is positive)
4. \dfrac{d^2C}{dx^2}=\dfrac{14400}{x^3}>0, so this is a minimum
5. C=2\times60+\dfrac{7200}{60}=120+120
6. Answer: \pounds240

Divide each term on the top by x first, so that y=x^2+128x^{-1}.

1. y=x^2+128x^{-1}
2. \dfrac{dy}{dx}=2x-128x^{-2}
3. 2x-\dfrac{128}{x^2}=0, so 2x^3=128
4. x^3=64, so x=4
5. \dfrac{d^2y}{dx^2}=2+\dfrac{256}{x^3}=6>0, so this is a minimum
6. y=16+\dfrac{128}{4}=16+32
7. Answer: 48

Solve \dfrac{dh}{dx}=0 and check which solution lies in the range 0\leqslant x\leqslant5.

1. \dfrac{dh}{dx}=6x^2-54x+84=6(x^2-9x+14)
2. 6(x-2)(x-7)=0, so x=2 or x=7
3. x=7 is outside 0\leqslant x\leqslant5, so use x=2
4. \dfrac{d^2h}{dx^2}=12x-54=-30<0, so this is a maximum
5. h=16-108+168+6=82
6. (At the ends, h=6 when x=0 and h=1 when x=5, both lower)
7. Answer: 82 m

Let B be the point (x,\ 0): then the rectangle is 2x wide and 12-x^2 high.

1. Let B=(x,\ 0), so C=(x,\ 12-x^2)
2. Width AB=2x, height BC=12-x^2
3. Area =2x(12-x^2)=24x-2x^3
4. \dfrac{d}{dx}(24x-2x^3)=24-6x^2=0, so x^2=4 and x=2
5. Second derivative -12x=-24<0, so this is a maximum
6. Area =4\times(12-4)=4\times8
7. Answer: 32

At the greatest yield \dfrac{dY}{dx}=0, so substitute x=7 into \dfrac{dY}{dx} to find k first.

1. \dfrac{dY}{dx}=k-6x
2. \dfrac{dY}{dx}=0 when x=7: k-42=0, so k=42
3. Y=40+42\times7-3\times7^2
4. =40+294-147
5. Answer: 187 tonnes

Write the number of people and the price of a ticket in terms of x, then multiply them to get R.

1. Number of people: 40+x; price of a ticket: 30-0.5x
2. R=(40+x)(30-0.5x)=1200+10x-0.5x^2
3. \dfrac{dR}{dx}=10-x=0, so x=10
4. \dfrac{d^2R}{dx^2}=-1<0, so this is a maximum
5. R=50\times25
6. Answer: \pounds1250

The rate of increase is \dfrac{dN}{dt}, so this time it is \dfrac{dN}{dt} that you need to maximise.

1. Rate of increase: \dfrac{dN}{dt}=60t-3t^2
2. To maximise this, differentiate again: \dfrac{d^2N}{dt^2}=60-6t
3. 60-6t=0, so t=10
4. The next derivative is -6<0, so this gives the greatest rate
5. \dfrac{dN}{dt}=60\times10-3\times10^2=600-300
6. Answer: 300 thousand downloads per week

Use the perimeter to write the arc length in terms of r; the arc length and the area are the same fraction \dfrac{\theta}{360} of the whole circle. Answer part (a) before attempting part (b).

Part (a)

1. The arc length is 20-2r
2. Arc length =\dfrac{\theta}{360}\times2\pi r, so \dfrac{\theta}{360}=\dfrac{20-2r}{2\pi r}
3. A=\dfrac{\theta}{360}\times\pi r^2=\dfrac{20-2r}{2\pi r}\times\pi r^2
4. A=\dfrac{r(20-2r)}{2}
5. Answer: A=10r-r^2

Part (b)

1. \dfrac{dA}{dr}=10-2r=0, so r=5
2. \dfrac{d^2A}{dr^2}=-2<0, so this is a maximum
3. A=50-25
4. Answer: 25 m^2

Part (c)

1. When r=5, the arc length is 20-10=10 m
2. \dfrac{\theta}{360}\times2\pi\times5=10
3. \theta=\dfrac{3600}{10\pi}=\dfrac{360}{\pi}=114.59\ldots
4. Answer: 114.6^\circ

Write the surface area as six faces to get h in terms of x, substitute into V=2x\times x\times h, then differentiate and set \dfrac{dV}{dx}=0.

Part (a)

1. Surface area: two faces 2x\times x, two faces 2x\times h, two faces x\times h
2. 2(2x^2)+2(2xh)+2(xh)=4x^2+6xh
3. 4x^2+6xh=432, so h=\dfrac{432-4x^2}{6x}
4. V=2x\times x\times h=2x^2\times\dfrac{432-4x^2}{6x}
5. V=\dfrac{x(432-4x^2)}{3}
6. Answer: V=144x-\frac43x^3

Part (b)

1. \dfrac{dV}{dx}=144-4x^2
2. Set 144-4x^2=0, so x^2=36 and x=6 (a length is positive)
3. \dfrac{d^2V}{dx^2}=-8x=-48<0, so this is a maximum
4. V=144\times6-\frac43\times6^3=864-288=576
5. Answer: 576 cm^3

Use the perimeter to write y in terms of x, and use \frac12ab\sin C for the area of the triangle. Answer part (a) before attempting part (b).

Part (a)

1. Perimeter: y+y+2x+2x+2x=18, so 2y+6x=18
2. y=9-3x
3. Rectangle: 2x(9-3x)=18x-6x^2
4. Triangle: \frac12\times2x\times2x\times\sin60^\circ=2x^2\times\dfrac{\sqrt3}{2}=\sqrt3x^2
5. A=18x-6x^2+\sqrt3x^2
6. Answer: A=18x-(6-\sqrt3)x^2

Part (b)

1. \dfrac{dA}{dx}=18-2(6-\sqrt3)x
2. \dfrac{dA}{dx}=0 when x=\dfrac{9}{6-\sqrt3}=2.108\ldots
3. \dfrac{d^2A}{dx^2}=-2(6-\sqrt3)<0, so this is a maximum
4. A=18\times2.108\ldots-(6-\sqrt3)\times2.108\ldots^2=18.97\ldots
5. Answer: 19.0 cm^2

Use the volume to write h in terms of r, then substitute into A=2\pi r^2+2\pi rh. Answer part (a) before attempting part (b).

Part (a)

1. Volume: \pi r^2h=500, so h=\dfrac{500}{\pi r^2}
2. A=2\pi r^2+2\pi rh (two ends and the curved surface)
3. 2\pi rh=2\pi r\times\dfrac{500}{\pi r^2}=\dfrac{1000}{r}
4. Answer: A=2\pi r^2+\dfrac{1000}{r}

Part (b)

1. A=2\pi r^2+1000r^{-1}
2. \dfrac{dA}{dr}=4\pi r-1000r^{-2}
3. 4\pi r=\dfrac{1000}{r^2}, so r^3=\dfrac{1000}{4\pi}=79.57\ldots
4. r=4.301\ldots
5. \dfrac{d^2A}{dr^2}=4\pi+\dfrac{2000}{r^3}>0, so this is a minimum
6. A=2\pi\times4.301\ldots^2+\dfrac{1000}{4.301\ldots}=116.2\ldots+232.4\ldots
7. Answer: 349 cm^2

Write P as (x,\ x^2) and use Pythagoras for AP^2; to solve the cubic, try small whole numbers and use the factor theorem. Answer part (a) before attempting part (b).

Part (a)

1. P=(x,\ x^2)
2. Horizontal distance x-3, vertical distance x^2
3. AP^2=(x-3)^2+(x^2)^2=x^2-6x+9+x^4
4. Answer: AP^2=x^4+x^2-6x+9

Part (b)

1. \dfrac{d}{dx}(x^4+x^2-6x+9)=4x^3+2x-6
2. Set it equal to 0 and divide by 2: 2x^3+x-3=0
3. x=1 works: 2+1-3=0, so (x-1) is a factor
4. 2x^3+x-3=(x-1)(2x^2+2x+3)
5. 2x^2+2x+3=0 has no solutions (discriminant 4-24<0)
6. Second derivative 12x^2+2=14>0, so this is a minimum
7. Answer: x=1

Part (c)

1. AP^2=1+1-6+9=5
2. Answer: AP=\sqrt5