Calculus BC · Unit 5
Turn a constraint into a function of one variable, then justify the maximum or minimum.
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One variable before you differentiate
Write the quantity you want to make large or small. Use the constraint to replace every other variable. Then differentiate.
Stay inside the domain that the story allows
A length has to be positive, and a cut square has to fit on the sheet. Check the critical points that survive, and check the ends of that domain.
Say why the critical point is a max or a min
Use the first derivative test, the second derivative test, or the values at the ends of a closed domain.
A square or a cube often wins
A rectangle of fixed perimeter with maximum area is a square. A closed box of fixed volume with minimum surface area, and a square base, comes out as a cube when the top is included.
A sum is fixed. Maximize a product, or minimize a sum of squares.
Let the numbers be \(x\) and \(20 - x\).
\(P = x(20 - x) = 20x - x^2\)
\(P' = 20 - 2x\)
\(x = 10\)
\(P'' = -2 < 0\), so this critical point is a maximum.
\(P = 100\)
Answer: 100
\(P = x(12 - x)\)
\(P' = 12 - 2x = 0\)
\(x = 6\)
\(P = 36\)
Answer: 36
\(S = x^2 + (30 - x)^2\)
\(S' = 2x + 2(30 - x)(-1) = 4x - 60\)
\(x = 15\)
\(S'' = 4 > 0\), so the critical point is a minimum.
\(S = 225 + 225 = 450\)
Answer: 450
\(2L + 2W = 40\), so \(W = 20 - L\).
\(A = L(20 - L)\)
\(A' = 20 - 2L = 0\)
\(L = 10\) and \(W = 10\)
\(A'' = -2 < 0\), a maximum. The rectangle is a square of area 100.
Answer: 10 by 10
\(R(p) = 40p - 2p^2\)
\(R' = 40 - 4p = 0\)
\(p = 10\)
\(R'' = -4 < 0\), a maximum.
Answer: \(p = 10\)
\(R(10) = 400 - 200\)
\(200\)
Answer: 200
A fence, a box, or a can. Use the constraint, then the first or second derivative test.
Let \(x\) be each side perpendicular to the barn, and let \(y\) be the side parallel to the barn.
\(2x + y = 60\), so \(y = 60 - 2x\).
\(A = x(60 - 2x) = 60x - 2x^2\)
\(A' = 60 - 4x = 0\)
\(x = 15\) and \(y = 30\)
\(A'' = -4 < 0\), a maximum. The area is 450 m².
Answer: 15 m by 30 m
\(x^2 h = 8\), so \(h = \dfrac{8}{x^2}\).
\(S = 2x^2 + 4xh = 2x^2 + \dfrac{32}{x}\)
\(S' = 4x - \dfrac{32}{x^2} = 0\)
\(4x^3 = 32\)
\(x = 2\) and \(h = 2\)
For \(0 < x < 2\), \(S' < 0\). For \(x > 2\), \(S' > 0\). The critical point is a minimum.
Answer: side 2
\(\pi r^2 h = 16\pi\), so \(h = \dfrac{16}{r^2}\).
\(S = 2\pi r^2 + \dfrac{32\pi}{r}\)
\(\dfrac{dS}{dr} = 4\pi r - \dfrac{32\pi}{r^2} = 0\)
\(r^3 = 8\)
\(r = 2\) and \(h = 4\)
Answer: \(r = 2\)
Put the base on the x-axis from \(-6\) to \(6\) and the top at \((0, 6)\).
The right edge of the triangle is \(y = 6 - x\) for \(0 \le x \le 6\).
A rectangle reaching that edge at horizontal distance \(x\) from the center has width \(2x\) and height \(6 - x\).
\(A = 2x(6 - x) = 12x - 2x^2\)
\(A' = 12 - 4x = 0\)
\(x = 3\), height 3, and \(A = 18\)
The area is 0 at the ends of the domain \(x = 0\) and \(x = 6\), so 18 is the maximum.
Answer: 18
At \(x = 2\) and \(h = 2\),
\(S = 2(4) + \dfrac{32}{2}\)
\(24\)
Answer: 24
\(x = 15\) and \(y = 30\) from \(2x + y = 60\).
Answer: A. 15 m by 30 m
A sheet of cardboard, a point on a line, and a window.
\(V = x(12 - 2x)^2\), with \(0 < x < 6\).
\(V' = (12 - 2x)^2 + x\cdot 2(12 - 2x)(-2)\)
\(V' = (12 - 2x)\bigl[(12 - 2x) - 4x\bigr]\)
\(V' = (12 - 2x)(12 - 6x) = 12(6 - x)(2 - x)\)
The critical point inside \((0, 6)\) is \(x = 2\).
\(V' > 0\) on \((0, 2)\) and \(V' < 0\) on \((2, 6)\), so the volume has a maximum.
\(V(2) = 2\cdot 8^2 = 128\)
Answer: \(x = 2\), volume 128
Minimize the square of the distance: \(D = x^2 + (2x + 1)^2\).
\(D = 5x^2 + 4x + 1\)
\(D' = 10x + 4 = 0\)
\(x = -\dfrac{2}{5}\)
\(y = 2\left(-\dfrac{2}{5}\right) + 1 = \dfrac{1}{5}\)
\(D'' = 10 > 0\), a minimum.
Distance \(\sqrt{\dfrac{4}{25} + \dfrac{1}{25}} = \sqrt{\dfrac{1}{5}} = \dfrac{\sqrt{5}}{5}\).
Answer: point \(\left(-\dfrac{2}{5}, \dfrac{1}{5}\right)\), distance \(\dfrac{\sqrt{5}}{5}\)
The width is \(2r\). Let \(h\) be the height of the rectangle.
\(2h + 2r + \pi r = 20\)
\(h = 10 - r - \dfrac{\pi r}{2}\)
\(A = 2rh + \dfrac{1}{2}\pi r^2\)
\(A = 20r - 2r^2 - \dfrac{\pi}{2} r^2\)
\(A' = 20 - 4r - \pi r = 0\)
\(r(4 + \pi) = 20\)
\(r = \dfrac{20}{4 + \pi}\)
Answer: \(r = \dfrac{20}{4 + \pi}\)
At \(x = 6\), the base of the box has side \(12 - 12 = 0\).
\(V(6) = 0\)
The sign of \(V'\) changes from positive to negative at \(x = 2\), and the volume there is 128.
Answer: the volume is 0 when \(x = 6\)
\(A' = 12 - 4x = 0\) at \(x = 3\).
\(A(3) = 36 - 18 = 18\)
\(A(0) = 0\) and \(A(6) = 0\)
Answer: 18
The minimum of \(x^2 + (2x + 1)^2\) occurs at \(x = -\dfrac{2}{5}\).
\(y = \dfrac{1}{5}\)
Answer: A. \(\left(-\dfrac{2}{5}, \dfrac{1}{5}\right)\)
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