Exercise 6.3 Practice
Maxima and Minima – NCERT Solutions
Similar to Q1
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Find the maximum and minimum values, if any, of the function given by $f(x) = (3x - 2)^2 + 5$.
Analysis: We know that $(3x - 2)^2 \ge 0$ for all $x \in \mathbb{R}$.
Minimum Value: The minimum value of $(3x - 2)^2$ is 0 (when $3x-2=0 \Rightarrow x=2/3$).
Therefore, minimum value of $f(x) = 0 + 5 = 5$.
Therefore, minimum value of $f(x) = 0 + 5 = 5$.
Maximum Value: As $x \to \infty$, $(3x-2)^2 \to \infty$. Thus, the function has no maximum value.
$\boxed{\text{Min Value: } 3, \text{ Max Value: None}}$
Similar to Q2
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Find the maximum and minimum values, if any, of the function given by $f(x) = -|x + 4| + 6$.
Analysis: We know that $|x + 4| \ge 0$ for all $x \in \mathbb{R}$.
$\Rightarrow -|x + 4| \le 0$.
$\Rightarrow -|x + 4| \le 0$.
Maximum Value: The maximum value of $-|x + 4|$ is 0 (when $x = -4$).
Therefore, maximum value of $f(x) = 0 + 6 = 6$.
Therefore, maximum value of $f(x) = 0 + 6 = 6$.
Minimum Value: As $x \to \pm\infty$, $-|x+4| \to -\infty$. Thus, no minimum value.
$\boxed{\text{Max Value: } 6, \text{ Min Value: None}}$
Similar to Q3
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Find the local maxima and local minima, if any, of the function $f(x) = x^3 - 3x$.
Derivative: $f'(x) = 3x^2 - 3$.
Critical Points: $3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Second Derivative Test: $f''(x) = 6x$.
At $x=1$: $f''(1) = 6 0 \Rightarrow$ Local Minima.
Value: $f(1) = 1 - 3 = -2$.
Value: $f(1) = 1 - 3 = -2$.
At $x=-1$: $f''(-1) = -6 0 \Rightarrow$ Local Maxima.
Value: $f(-1) = -1 + 3 = 2$.
Value: $f(-1) = -1 + 3 = 2$.
$\boxed{\text{Local Max: } 2 \text{ at } x=-1, \text{ Local Min: } -2 \text{ at } x=1}$
Similar to Q4
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Prove that the function $f(x) = e^{2x}$ does not have maxima or minima.
Derivative: $f'(x) = 2e^{2x}$.
Analysis: We know that exponential function $e^{2x} 0$ for all $x \in \mathbb{R}$.
Therefore, $f'(x) \neq 0$ for any real $x$.
Since there are no critical points, the function has no local maxima or minima.
$\boxed{\text{Proven}}$
Similar to Q5
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Find the absolute maximum value and the absolute minimum value of the function $f(x) = x^3 - 6x^2 + 9x + 15$ on the interval $[0, 5]$.
Derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$.
Critical Points: $x = 1, x = 3$. Both in $[0, 5]$.
Evaluate at Critical Points & Endpoints:
$f(0) = 15$
$f(1) = 1 - 6 + 9 + 15 = 19$
$f(3) = 27 - 54 + 27 + 15 = 15$
$f(5) = 125 - 150 + 45 + 15 = 35$
$f(0) = 15$
$f(1) = 1 - 6 + 9 + 15 = 19$
$f(3) = 27 - 54 + 27 + 15 = 15$
$f(5) = 125 - 150 + 45 + 15 = 35$
$\boxed{\text{Abs Max: } 35, \text{ Abs Min: } 15}$
Similar to Q6
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Find the maximum profit that a company can make, if the profit function is given by $P(x) = 50 - 20x - 5x^2$.
Derivative: $P'(x) = -20 - 10x$.
Critical Point: $-20 - 10x = 0 \Rightarrow 10x = -20 \Rightarrow x = -2$.
Second Derivative: $P''(x) = -10 0$ (Maxima).
Max Profit: $P(-2) = 50 - 20(-2) - 5(-2)^2 = 50 + 40 - 20 = 70$.
$\boxed{70}$
Similar to Q7
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Find both the maximum value and the minimum value of $3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval $[0, 3]$.
Derivative: $f'(x) = 12x^3 - 24x^2 + 24x - 48 = 12(x^3 - 2x^2 + 2x - 4)$.
Factorize: $12[x^2(x-2) + 2(x-2)] = 12(x-2)(x^2+2)$.
Critical Point: $x = 2$ (since $x^2+2 \neq 0$).
Evaluate:
$f(0) = 25$
$f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39$
$f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16$
$f(0) = 25$
$f(2) = 3(16) - 8(8) + 12(4) - 48(2) + 25 = 48 - 64 + 48 - 96 + 25 = -39$
$f(3) = 3(81) - 8(27) + 12(9) - 48(3) + 25 = 243 - 216 + 108 - 144 + 25 = 16$
$\boxed{\text{Max: } 25, \text{ Min: } -39}$
Similar to Q8
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At what points in the interval $[0, 2\pi]$, does the function $\sin 2x$ attain its maximum value?
Derivative: $f(x) = \sin 2x \Rightarrow f'(x) = 2\cos 2x$.
Critical Points: $2\cos 2x = 0 \Rightarrow 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$.
$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
Second Derivative: $f''(x) = -4\sin 2x$.
Check Maxima ($f'' 0$):
At $x=\frac{\pi}{4}: -4\sin(\frac{\pi}{2}) = -4 0$ (Max)
At $x=\frac{5\pi}{4}: -4\sin(\frac{5\pi}{2}) = -4 0$ (Max)
At $x=\frac{\pi}{4}: -4\sin(\frac{\pi}{2}) = -4 0$ (Max)
At $x=\frac{5\pi}{4}: -4\sin(\frac{5\pi}{2}) = -4 0$ (Max)
$\boxed{x = \frac{\pi}{4}, \frac{5\pi}{4}}$
Similar to Q9
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What is the maximum value of the function $\sin x + \cos x$?
Method 1 (Calculus): $f'(x) = \cos x - \sin x = 0 \Rightarrow \tan x = 1 \Rightarrow x = \pi/4$.
$f(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
$f(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
Method 2 (Trigonometry): $\sin x + \cos x = \sqrt{2}(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x) = \sqrt{2}\sin(x + \frac{\pi}{4})$.
Max value of sine is 1, so max value is $\sqrt{2}(1) = \sqrt{2}$.
Max value of sine is 1, so max value is $\sqrt{2}(1) = \sqrt{2}$.
$\boxed{\sqrt{2}}$
Similar to Q10
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Find the maximum value of $2x^3 - 24x + 107$ in the interval $[1, 3]$. Find the maximum value of the same function in $[-3, -1]$.
Derivative: $f'(x) = 6x^2 - 24 = 6(x^2 - 4)$. Critical points $x = \pm 2$.
Interval $[1, 3]$: Critical point $x=2$.
$f(1) = 2 - 24 + 107 = 85$
$f(2) = 16 - 48 + 107 = 75$
$f(3) = 54 - 72 + 107 = 89$
Max Value: 89.
$f(1) = 2 - 24 + 107 = 85$
$f(2) = 16 - 48 + 107 = 75$
$f(3) = 54 - 72 + 107 = 89$
Max Value: 89.
Interval $[-3, -1]$: Critical point $x=-2$.
$f(-3) = -54 + 72 + 107 = 125$
$f(-2) = -16 + 48 + 107 = 139$
$f(-1) = -2 + 24 + 107 = 129$
Max Value: 139.
$f(-3) = -54 + 72 + 107 = 125$
$f(-2) = -16 + 48 + 107 = 139$
$f(-1) = -2 + 24 + 107 = 129$
Max Value: 139.
$\boxed{89 \text{ and } 139}$
Similar to Q11
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It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$. Find the value of $a$.
Condition: Since max is at $x=1$, $f'(1) = 0$.
Derivative: $f'(x) = 4x^3 - 124x + a$.
Substitute $x=1$: $4(1)^3 - 124(1) + a = 0$.
$4 - 124 + a = 0 \Rightarrow a = 120$.
$4 - 124 + a = 0 \Rightarrow a = 120$.
$\boxed{120}$
Similar to Q12
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Find the maximum and minimum values of $x + \sin 2x$ on $[0, 2\pi]$.
Derivative: $f'(x) = 1 + 2\cos 2x$.
Critical Points: $\cos 2x = -1/2$.
$2x = 2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3$.
$x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$.
$2x = 2\pi/3, 4\pi/3, 8\pi/3, 10\pi/3$.
$x = \pi/3, 2\pi/3, 4\pi/3, 5\pi/3$.
Evaluate:
$f(0) = 0$
$f(\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.91$
$f(2\pi/3) = 2\pi/3 - \sqrt{3}/2 \approx 1.22$
$f(4\pi/3) = 4\pi/3 + \sqrt{3}/2 \approx 5.05$
$f(5\pi/3) = 5\pi/3 - \sqrt{3}/2 \approx 4.36$
$f(2\pi) = 2\pi \approx 6.28$
$f(0) = 0$
$f(\pi/3) = \pi/3 + \sqrt{3}/2 \approx 1.91$
$f(2\pi/3) = 2\pi/3 - \sqrt{3}/2 \approx 1.22$
$f(4\pi/3) = 4\pi/3 + \sqrt{3}/2 \approx 5.05$
$f(5\pi/3) = 5\pi/3 - \sqrt{3}/2 \approx 4.36$
$f(2\pi) = 2\pi \approx 6.28$
$\boxed{\text{Max: } 2\pi, \text{ Min: } 0}$
Similar to Q13
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Find two numbers whose sum is 24 and whose product is as large as possible.
Let numbers be $x, y$: $x+y=24 \Rightarrow y=24-x$.
Maximize $P = xy$: $P(x) = x(24-x) = 24x - x^2$.
Derivative: $P'(x) = 24 - 2x = 0 \Rightarrow x = 12$.
Second Derivative: $P''(x) = -20$ (Maxima).
Numbers: $x=12, y=12$.
$\boxed{12, 12}$
Similar to Q14
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Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.
Relation: $x = 60 - y$.
Function $P(y) = (60-y)y^3 = 60y^3 - y^4$.
Derivative: $P'(y) = 180y^2 - 4y^3 = 4y^2(45 - y)$.
Critical Point: $y = 45$ (since $y0$).
Second Derivative: $P''(y) = 360y - 12y^2$.
At $y=45$: $P''(45) = 12(45)(30 - 45) 0$ (Maxima).
At $y=45$: $P''(45) = 12(45)(30 - 45) 0$ (Maxima).
Values: $y=45, x=15$.
$\boxed{x=15, y=45}$
Similar to Q15
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Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^2 y^5$ is a maximum.
Relation: $x + y = 35 \Rightarrow x = 35 - y$.
Function $P(y) = (35-y)^2 y^5$.
Derivative: $P'(y) = 2(35-y)(-1)y^5 + (35-y)^2(5y^4)$
$= y^4(35-y) [-2y + 5(35-y)] = y^4(35-y)(175 - 7y)$.
$= y^4(35-y) [-2y + 5(35-y)] = y^4(35-y)(175 - 7y)$.
Critical Point: $175 - 7y = 0 \Rightarrow y = 25$.
Check Max: At $y=25$, sign changes from + to -. Maxima.
Values: $y=25, x=10$.
$\boxed{x=10, y=25}$
Similar to Q16
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Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Relation: $y = 16 - x$.
Function $S(x) = x^3 + (16-x)^3$.
Derivative: $S'(x) = 3x^2 + 3(16-x)^2(-1) = 3[x^2 - (16-x)^2]$.
Critical Point: $x^2 = (16-x)^2 \Rightarrow x = 16-x \Rightarrow 2x = 16 \Rightarrow x = 8$.
Second Derivative: $S''(x) = 6x + 6(16-x) = 96 0$ (Minima).
$\boxed{8, 8}$
Similar to Q17
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A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Volume: $V(x) = x(18-2x)^2$.
Derivative: $V'(x) = (18-2x)^2 - 4x(18-2x) = (18-2x)(18-6x)$.
Critical Points: $x=9$ (not possible) or $x=3$.
Second Derivative: $V''(x) = -12(18-2x) - 2(18-6x)$.
At $x=3$: $V''(3) = -12(12) - 0 = -144 0$ (Maxima).
At $x=3$: $V''(3) = -12(12) - 0 = -144 0$ (Maxima).
$\boxed{3 \text{ cm}}$
Similar to Q18
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A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Volume: $V(x) = x(45-2x)(24-2x) = 4x(45-2x)(12-x)$.
Simplify: $V(x) = 4(540x - 69x^2 + 2x^3)$.
Derivative: $V'(x) = 4(540 - 138x + 6x^2) = 24(x^2 - 23x + 90)$.
Factors: $24(x-18)(x-5) = 0$.
Critical Points: $x=18$ (not possible as width is 24) or $x=5$.
Check Max: $V''(5) 0$.
$\boxed{5 \text{ cm}}$
Similar to Q19
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Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Relation: $x^2 + y^2 = (2R)^2 \Rightarrow y = \sqrt{4R^2 - x^2}$.
Area Squared: $Z = A^2 = x^2(4R^2 - x^2) = 4R^2x^2 - x^4$.
Derivative: $Z' = 8R^2x - 4x^3 = 0 \Rightarrow x^2 = 2R^2 \Rightarrow x = R\sqrt{2}$.
Find y: $y = \sqrt{4R^2 - 2R^2} = R\sqrt{2}$.
Conclusion: $x = y$, so it is a square.
$\boxed{\text{Proven}}$
Similar to Q20
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Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Surface Area (S): $S = 2\pi r^2 + 2\pi rh \Rightarrow h = \frac{S - 2\pi r^2}{2\pi r}$.
Volume (V): $V = \pi r^2 h = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = \frac{r}{2}(S - 2\pi r^2) = \frac{Sr}{2} - \pi r^3$.
Derivative: $\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 = 0 \Rightarrow S = 6\pi r^2$.
Substitute S: $2\pi r^2 + 2\pi rh = 6\pi r^2 \Rightarrow 2\pi rh = 4\pi r^2 \Rightarrow h = 2r$.
Conclusion: Height = Diameter.
$\boxed{\text{Proven}}$
Similar to Q21
00:00
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Volume: $\pi r^2 h = 100 \Rightarrow h = \frac{100}{\pi r^2}$.
Surface Area: $S = 2\pi r^2 + 2\pi r \left(\frac{100}{\pi r^2}\right) = 2\pi r^2 + \frac{200}{r}$.
Derivative: $\frac{dS}{dr} = 4\pi r - \frac{200}{r^2} = 0 \Rightarrow 4\pi r^3 = 200 \Rightarrow r^3 = \frac{50}{\pi}$.
Radius: $r = \sqrt[3]{\frac{50}{\pi}}$.
Height: $h = 2r$ (from optimization condition).
$\boxed{r = \sqrt[3]{\frac{50}{\pi}}, h = 2\sqrt[3]{\frac{50}{\pi}}}$
Similar to Q22
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A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Let pieces be $x$ (circle) and $28-x$ (square).
Circle: $2\pi r = x \Rightarrow r = \frac{x}{2\pi}$. Area $A_c = \frac{x^2}{4\pi}$.
Square: $4s = 28-x \Rightarrow s = \frac{28-x}{4}$. Area $A_s = \frac{(28-x)^2}{16}$.
Total Area A: $A = \frac{x^2}{4\pi} + \frac{(28-x)^2}{16}$.
Derivative: $A' = \frac{2x}{4\pi} + \frac{2(28-x)(-1)}{16} = \frac{x}{2\pi} - \frac{28-x}{8}$.
Critical Point: $\frac{x}{2\pi} = \frac{28-x}{8} \Rightarrow 4x = \pi(28-x) \Rightarrow x(4+\pi) = 28\pi$.
Lengths: $x = \frac{28\pi}{4+\pi}$ m (Circle), Square piece = $28 - x = \frac{112}{4+\pi}$ m.
$\boxed{\text{Circle: } \frac{28\pi}{4+\pi}, \text{ Square: } \frac{112}{4+\pi}}$
Similar to Q23
00:00
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.
Setup: Let cone height be $h$ and radius $r$. Let sphere radius be $R$.
$h = R + x$ where $x$ is distance from center. $r^2 = R^2 - x^2$.
$h = R + x$ where $x$ is distance from center. $r^2 = R^2 - x^2$.
Volume V: $V = \frac{1}{3}\pi (R^2 - x^2)(R+x)$.
Maximize V: $\frac{dV}{dx} = \frac{\pi}{3} [ (R^2-x^2)(1) + (R+x)(-2x) ] = \frac{\pi}{3}(R+x)(R-x-2x) = \frac{\pi}{3}(R+x)(R-3x)$.
Critical Point: $x = R/3$. So $h = R + R/3 = 4R/3$.
Max Volume: $V = \frac{1}{3}\pi (R^2 - R^2/9)(4R/3) = \frac{1}{3}\pi (\frac{8R^2}{9})(\frac{4R}{3}) = \frac{32\pi R^3}{81}$.
Ratio: $\frac{V}{V_{sphere}} = \frac{32\pi R^3/81}{4\pi R^3/3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}$.
$\boxed{\text{Proven}}$
Similar to Q24
00:00
Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ time the radius of the base.
Volume V: $V = \frac{1}{3}\pi r^2 h \Rightarrow h = \frac{3V}{\pi r^2}$.
Curved Surface S: $S = \pi r l = \pi r \sqrt{r^2 + h^2}$.
Minimize $S^2$: $Z = S^2 = \pi^2 r^2 (r^2 + \frac{9V^2}{\pi^2 r^4}) = \pi^2 r^4 + \frac{9V^2}{r^2}$.
Derivative: $Z' = 4\pi^2 r^3 - \frac{18V^2}{r^3} = 0 \Rightarrow 2\pi^2 r^6 = 9V^2$.
Substitute V: $9(\frac{1}{9}\pi^2 r^4 h^2) = 2\pi^2 r^6 \Rightarrow \pi^2 r^4 h^2 = 2\pi^2 r^6 \Rightarrow h^2 = 2r^2 \Rightarrow h = \sqrt{2}r$.
$\boxed{\text{Proven}}$
Similar to Q25
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Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.
Variables: Slant height $l$ (constant), height $h$, radius $r$. $r^2 = l^2 - h^2$.
Volume V: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l^2 - h^2)h = \frac{1}{3}\pi (l^2h - h^3)$.
Derivative: $\frac{dV}{dh} = \frac{\pi}{3}(l^2 - 3h^2) = 0 \Rightarrow l^2 = 3h^2 \Rightarrow l = \sqrt{3}h$.
Angle $\alpha$: $\cos \alpha = \frac{h}{l} = \frac{1}{\sqrt{3}}$.
Find $\tan \alpha$: $\tan^2 \alpha = \sec^2 \alpha - 1 = 3 - 1 = 2 \Rightarrow \tan \alpha = \sqrt{2}$.
$\boxed{\alpha = \tan^{-1}\sqrt{2}}$
Similar to Q26
00:00
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin^{-1}(\frac{1}{3})$.
Surface Area S: $S = \pi r^2 + \pi r l \Rightarrow l = \frac{S - \pi r^2}{\pi r}$.
Volume Squared $Z$: $Z = V^2 = \frac{1}{9}\pi^2 r^4 (l^2 - r^2)$. Substitute $l$.
Simplify: After differentiation and solving, we get $S = 4\pi r^2$.
Relation: $4\pi r^2 = \pi r^2 + \pi r l \Rightarrow 3\pi r^2 = \pi r l \Rightarrow l = 3r$.
Angle $\alpha$: $\sin \alpha = \frac{r}{l} = \frac{r}{3r} = \frac{1}{3}$.
$\boxed{\alpha = \sin^{-1}(\frac{1}{3})}$
Similar to Q27 (MCQ)
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The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is:
(A) $(2\sqrt{2}, 4)$
(B) $(2\sqrt{2}, 0)$
(C) $(0, 0)$
(D) $(2, 2)$
(A) $(2\sqrt{2}, 4)$
(B) $(2\sqrt{2}, 0)$
(C) $(0, 0)$
(D) $(2, 2)$
Distance Squared $D$: $D = x^2 + (y-5)^2 = 2y + (y-5)^2$.
Derivative: $D'(y) = 2 + 2(y-5) = 2y - 8$.
Critical Point: $2y - 8 = 0 \Rightarrow y = 4$.
Find x: $x^2 = 2(4) = 8 \Rightarrow x = \pm 2\sqrt{2}$.
Point: $(2\sqrt{2}, 4)$.
$\boxed{\text{(A) } (2\sqrt{2}, 4)}$
Similar to Q28 (MCQ)
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For all real values of $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is:
(A) 0
(B) 1
(C) 3
(D) 1/3
(A) 0
(B) 1
(C) 3
(D) 1/3
Let $y = \frac{1-x+x^2}{1+x+x^2}$.
Derivative: $y' = \frac{(2x-1)(1+x+x^2) - (1-x+x^2)(2x+1)}{(1+x+x^2)^2}$.
Simplify Numerator: $2(x^2-1)$.
Critical Points: $x = \pm 1$.
Values: $y(1) = 1/3$, $y(-1) = 3$. Min is 1/3.
$\boxed{\text{(D) } 1/3}$
Similar to Q29 (MCQ)
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The maximum value of $[x(x-1) + 1]^{\frac{1}{3}}, 0 \le x \le 1$ is:
(A) $(1/3)^{1/3}$
(B) 1/2
(C) 1
(D) 0
(A) $(1/3)^{1/3}$
(B) 1/2
(C) 1
(D) 0
Let $g(x) = x^2 - x + 1$. Maximize $g(x)$ on $[0, 1]$.
Endpoints: $g(0) = 1$, $g(1) = 1$.
Vertex: $x = 1/2, g(1/2) = 3/4$ (Min).
Max of g(x) is 1. So max value is $1^{1/3} = 1$.
$\boxed{\text{(C) } 1}$