Question 1

A rectangular box with an open top is to be constructed from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume of the box as a function of x is V(x) = x(12 - 2x)². For what value of x is the volume maximum?

Accepted Answer

To find the maximum volume, we use the first derivative test.

Given: V(x) = x(12 - 2x)²

Step 1: Expand the function.
V(x) = x(144 - 48x + 4x²) = 144x - 48x² + 4x³

Step 2: Find the first derivative.
dV/dx = 144 - 96x + 12x²

Step 3: Set dV/dx = 0 to find critical points.
12x² - 96x + 144 = 0

Step 4: Divide by 12.
x² - 8x + 12 = 0

Step 5: Factor the quadratic.
(x - 2)(x - 6) = 0

Step 6: Solve for x.
x = 2 or x = 6

Step 7: Check validity. Since the sheet is 12 cm × 12 cm and we cut x from each corner, we need 0 < x < 6. So x = 6 is not valid (it would result in zero dimensions).

Step 8: Verify x = 2 is a maximum using the second derivative test.
d²V/dx² = -96 + 24x
At x = 2: d²V/dx² = -96 + 48 = -48 < 0 ✓ (confirms maximum)

Therefore, x = 2 cm gives the maximum volume.

Question 2

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Accepted Answer

This is a related rates problem. We use implicit differentiation with the Pythagorean theorem.

Step 1: Set up the relationship.
Let x = distance of the bottom from the wall (in metres)
Let y = height of the top on the wall (in metres)
Ladder length: x² + y² = 10² = 100 (constant)

Step 2: Differentiate both sides with respect to time t.
d/dt(x² + y²) = d/dt(100)
2x(dx/dt) + 2y(dy/dt) = 0

Step 3: Simplify.
x(dx/dt) + y(dy/dt) = 0

Step 4: Identify known values.
dx/dt = 2 m/s (bottom slides away)
When x = 6 m, find y:
6² + y² = 100
36 + y² = 100
y² = 64
y = 8 m (taking positive value)

Step 5: Solve for dy/dt.
6(2) + 8(dy/dt) = 0
12 + 8(dy/dt) = 0
dy/dt = -12/8 = -3/2 = -1.5 m/s

The negative sign indicates the top is moving downward.
The speed is 1.5 m/s downward, or |dy/dt| = 1.5 m/s.

Question 3

A function f(x) = x⁴ - 8x² + 5 has local extrema at x = a and x = b. The sum a + b equals:

Accepted Answer

To find local extrema, we set the first derivative equal to zero.

Step 1: Differentiate the function.
f(x) = x⁴ - 8x² + 5
f'(x) = 4x³ - 16x

Step 2: Set f'(x) = 0.
4x³ - 16x = 0
4x(x² - 4) = 0
4x(x - 2)(x + 2) = 0

Step 3: Solve for critical points.
x = 0, x = 2, or x = -2

Step 4: Determine which are local extrema using the second derivative test.
f''(x) = 12x² - 16

At x = 0: f''(0) = 0 - 16 = -16 < 0 (local maximum)
At x = 2: f''(2) = 12(4) - 16 = 48 - 16 = 32 > 0 (local minimum)
At x = -2: f''(-2) = 12(4) - 16 = 48 - 16 = 32 > 0 (local minimum)

Step 5: Identify a and b as the two local minima.
a = -2, b = 2 (or vice versa)
a + b = -2 + 2 = 0

Note: x = 0 is a local maximum, not a local extremum in the sense of minima, but it is a local extremum overall. However, the question asks for local extrema at x = a and x = b. If we interpret this as the two distinct local minima: a + b = 0. If we include all three critical points where local extrema occur, we still have the sum of the two minima as 0.

Question 4

The equation of the tangent line to the curve y = x³ - 3x² + 2 at the point where x = 1 is:

Accepted Answer

To find the equation of the tangent line, we need the slope at x = 1 and the point on the curve.

Step 1: Find the y-coordinate at x = 1.
y(1) = (1)³ - 3(1)² + 2 = 1 - 3 + 2 = 0
Point: (1, 0)

Step 2: Find the slope by differentiating.
dy/dx = 3x² - 6x

Step 3: Evaluate the derivative at x = 1.
dy/dx|ₓ₌₁ = 3(1)² - 6(1) = 3 - 6 = -3
Slope m = -3

Step 4: Use point-slope form of a line.
y - y₁ = m(x - x₁)
y - 0 = -3(x - 1)
y = -3x + 3

Alternatively in standard form: 3x + y - 3 = 0

The equation of the tangent line is y = -3x + 3.

Question 5

A rectangular box with an open top is to be constructed from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume of the box is maximized when x equals:

Accepted Answer

We need to find the value of x that maximizes the volume of the box. Step 1: Set up the volume function. After cutting squares of side x from each corner and folding: - Base dimensions: (12 - 2x) × (12 - 2x) - Height: x - Volume: V(x) = x(12 - 2x)² = x(144 - 48x + 4x²) = 144x - 48x² + 4x³ Step 2: Find the critical points by differentiating. dV/dx = 144 - 96x + 12x² Step 3: Set dV/dx = 0. 12x² - 96x + 144 = 0 Divide by 12: x² - 8x + 12 = 0 Step 4: Factor or use quadratic formula. (x - 2)(x - 6) = 0 x = 2 or x = 6 Step 5: Apply physical constraints and second derivative test. For a valid box: 0 < x < 6 (since 12 - 2x > 0) At x = 6: base dimensions become 0, so this is invalid. Second derivative: d²V/dx² = 24x - 96 At x = 2: d²V/dx² = 48 - 96 = -48 < 0 (maximum) At x = 6: d²V/dx² = 144 - 96 = 48 > 0 (minimum, but invalid anyway) Therefore, x = 2 cm maximizes the volume.

Question 6

A particle moves along a straight line such that its position at time t is given by s(t) = t³ - 6t² + 9t + 2 metres. At what time(s) is the particle momentarily at rest?

Accepted Answer

The particle is at rest when its velocity is zero. Velocity is the first derivative of position.

Given: s(t) = t³ - 6t² + 9t + 2

Step 1: Find velocity by differentiating position.
v(t) = ds/dt = 3t² - 12t + 9

Step 2: Set velocity equal to zero.
3t² - 12t + 9 = 0

Step 3: Simplify by dividing by 3.
t² - 4t + 3 = 0

Step 4: Factor the quadratic.
(t - 1)(t - 3) = 0

Step 5: Solve for t.
t = 1 s or t = 3 s

The particle is momentarily at rest at t = 1 second and t = 3 seconds.

Question 7

A rectangular box with an open top is to be constructed from a square piece of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume V of the resulting box as a function of x is V(x) = x(12 - 2x)². For what value of x is the volume maximum?

Accepted Answer

To find the maximum volume, we use the first derivative test.

Given: V(x) = x(12 - 2x)²

Step 1: Expand the function.
V(x) = x(144 - 48x + 4x²) = 144x - 48x² + 4x³

Step 2: Differentiate with respect to x.
dV/dx = 144 - 96x + 12x²

Step 3: Set dV/dx = 0 to find critical points.
12x² - 96x + 144 = 0

Step 4: Divide by 12.
x² - 8x + 12 = 0

Step 5: Factor the quadratic.
(x - 2)(x - 6) = 0

Step 6: Find roots.
x = 2 or x = 6

Step 7: Check validity. Since the cardboard side is 12 cm, we need 2x < 12, so x < 6. Thus x = 6 is invalid (it would result in zero dimensions).

Step 8: Verify x = 2 is a maximum using the second derivative test.
d²V/dx² = -96 + 24x
At x = 2: d²V/dx² = -96 + 48 = -48 < 0 ✓ (confirms maximum)

The volume is maximum when x = 2 cm.

Question 8

The equation of a curve is y = x³ - 3x² + 2x + 5. At what point on the curve is the tangent line parallel to the x-axis?

Accepted Answer

A tangent line parallel to the x-axis has slope 0. The slope of the tangent at any point is given by dy/dx.

Given: y = x³ - 3x² + 2x + 5

Step 1: Differentiate to find the slope of the tangent.
dy/dx = 3x² - 6x + 2

Step 2: Set dy/dx = 0 for tangent parallel to x-axis.
3x² - 6x + 2 = 0

Step 3: Use the quadratic formula.
x = [6 ± √(36 - 24)] / 6 = [6 ± √12] / 6 = [6 ± 2√3] / 6 = (3 ± √3) / 3

Step 4: Simplify.
x₁ = (3 - √3) / 3 ≈ 0.423
x₂ = (3 + √3) / 3 ≈ 1.577

Step 5: Find corresponding y-values by substituting into the original equation.
For x = (3 - √3) / 3: y = [(3 - √3) / 3]³ - 3[(3 - √3) / 3]² + 2[(3 - √3) / 3] + 5

For x = (3 + √3) / 3: y = [(3 + √3) / 3]³ - 3[(3 + √3) / 3]² + 2[(3 + √3) / 3] + 5

The tangent line is parallel to the x-axis at two points with x-coordinates x = (3 - √3) / 3 and x = (3 + √3) / 3.

Question 9

A function f(x) = x⁴ - 4x³ + 6x² - 4x + 1 has a critical point at x = 1. What is the nature of this critical point?

Accepted Answer

To determine the nature of a critical point, we use the second derivative test.

Given: f(x) = x⁴ - 4x³ + 6x² - 4x + 1

Step 1: Find the first derivative.
f'(x) = 4x³ - 12x² + 12x - 4

Step 2: Verify x = 1 is a critical point.
f'(1) = 4(1) - 12(1) + 12(1) - 4 = 4 - 12 + 12 - 4 = 0 ✓

Step 3: Find the second derivative.
f''(x) = 12x² - 24x + 12

Step 4: Evaluate the second derivative at x = 1.
f''(1) = 12(1)² - 24(1) + 12 = 12 - 24 + 12 = 0

Step 5: Since f''(1) = 0, the second derivative test is inconclusive. Use the third derivative test or analyze the sign of f'(x) around x = 1.

f'''(x) = 24x - 24
f'''(1) = 24(1) - 24 = 0

Step 6: Since f'''(1) = 0 as well, check the fourth derivative.
f⁽⁴⁾(x) = 24
f⁽⁴⁾(1) = 24 > 0

Step 7: Since the first non-zero derivative at x = 1 is the fourth derivative (even order) and it is positive, x = 1 is a point of local minimum.

Alternatively, note that f(x) = (x - 1)⁴, so x = 1 is a local (and global) minimum.

Question 10

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder is being pulled away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Accepted Answer

This is a related rates problem. We use implicit differentiation with the Pythagorean theorem.

Given:
- Ladder length L = 10 m (constant)
- Rate at which bottom moves away: dx/dt = 2 m/s (constant)
- Find: dy/dt when x = 6 m

Step 1: Set up the relationship using the Pythagorean theorem.
Let x = distance of bottom from wall, y = height of top on wall.
x² + y² = 10² = 100

Step 2: Differentiate both sides with respect to time t.
2x(dx/dt) + 2y(dy/dt) = 0

Step 3: Simplify.
x(dx/dt) + y(dy/dt) = 0

Step 4: Solve for dy/dt.
dy/dt = -x(dx/dt) / y

Step 5: Find y when x = 6.
6² + y² = 100
36 + y² = 100
y² = 64
y = 8 m (taking positive value)

Step 6: Substitute known values.
dy/dt = -(6)(2) / 8 = -12/8 = -3/2 = -1.5 m/s

The negative sign indicates the top is moving downward. The speed is 1.5 m/s downward.

Question 11

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 1 m/s. When the bottom is 6 m away from the wall, at what rate is the top of the ladder sliding down the wall?

Accepted Answer

This is a related rates problem involving the Pythagorean theorem.

Let x = distance of the bottom of the ladder from the wall (in metres)
Let y = height of the top of the ladder on the wall (in metres)
Given: ladder length = 10 m (constant), dx/dt = 1 m/s

Step 1: Set up the constraint using the Pythagorean theorem.
x² + y² = 10² = 100

Step 2: Differentiate both sides with respect to time t.
2x(dx/dt) + 2y(dy/dt) = 0

Step 3: Simplify.
x(dx/dt) + y(dy/dt) = 0

Step 4: Solve for dy/dt.
dy/dt = -x(dx/dt)/y

Step 5: Find y when x = 6.
From x² + y² = 100:
6² + y² = 100
36 + y² = 100
y² = 64
y = 8 m (taking positive value)

Step 6: Substitute known values into the rate equation.
dy/dt = -(6)(1)/8 = -6/8 = -3/4 m/s

The negative sign indicates the top is sliding down. The magnitude of the rate is 3/4 m/s or 0.75 m/s.

Question 12

A function f(x) = x³ - 3x² - 9x + 5 has local extrema. Which of the following correctly identifies the nature of the critical points?

Accepted Answer

To classify critical points, we find where f'(x) = 0 and use the second derivative test.

Given: f(x) = x³ - 3x² - 9x + 5

Step 1: Find the first derivative.
f'(x) = 3x² - 6x - 9

Step 2: Set f'(x) = 0 to find critical points.
3x² - 6x - 9 = 0

Step 3: Divide by 3.
x² - 2x - 3 = 0

Step 4: Factor.
(x - 3)(x + 1) = 0

Step 5: Solve for critical points.
x = 3 or x = -1

Step 6: Find the second derivative.
f''(x) = 6x - 6

Step 7: Apply the second derivative test.
At x = -1:
f''(-1) = 6(-1) - 6 = -12 < 0 → local maximum

At x = 3:
f''(3) = 6(3) - 6 = 12 > 0 → local minimum

Therefore, x = -1 is a local maximum and x = 3 is a local minimum.

Question 13

The radius of a circular disc is increasing at a constant rate of 2 cm/s. At the instant when the radius is 5 cm, what is the rate of change of the area of the disc?

Accepted Answer

This is a related rates problem using the chain rule.

Given information:
- Radius r is increasing at dr/dt = 2 cm/s
- We need to find dA/dt when r = 5 cm
- Area of a circle: A = πr²

Step 1: Differentiate the area formula with respect to time t.
dA/dt = d/dt(πr²) = π · 2r · dr/dt (by chain rule)

Step 2: Substitute the known values.
At r = 5 cm and dr/dt = 2 cm/s:
dA/dt = π · 2(5) · 2
dA/dt = π · 10 · 2
dA/dt = 20π cm²/s

Therefore, the area is increasing at a rate of 20π cm²/s.

Question 14

A rectangular box with an open top is to be made from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume V of the box as a function of x is V(x) = x(12 − 2x)². For what value of x is the volume maximum?

Accepted Answer

To find the maximum volume, we differentiate V(x) = x(12 − 2x)² and set dV/dx = 0.

Using the product rule: dV/dx = (12 − 2x)² + x · d/dx[(12 − 2x)²]

For d/dx[(12 − 2x)²], use the chain rule:
d/dx[(12 − 2x)²] = 2(12 − 2x) · (−2) = −4(12 − 2x)

Therefore:
dV/dx = (12 − 2x)² + x · [−4(12 − 2x)]
dV/dx = (12 − 2x)² − 4x(12 − 2x)
dV/dx = (12 − 2x)[(12 − 2x) − 4x]
dV/dx = (12 − 2x)(12 − 6x)

Setting dV/dx = 0:
(12 − 2x)(12 − 6x) = 0

This gives x = 6 or x = 2.

Since x must satisfy 0 < x < 6 (the side of the cut square must be less than half the cardboard side), we check both critical points and the boundary:
- At x = 2: V(2) = 2(12 − 4)² = 2(64) = 128 cm³
- At x = 6: V(6) = 6(12 − 12)² = 0 cm³

Therefore, the maximum volume occurs at x = 2 cm.

Question 15

A particle moves along a curve y = x³ − 3x² + 2x. At what point(s) on the curve is the tangent line horizontal?

Accepted Answer

A horizontal tangent occurs when dy/dx = 0. First, differentiate y = x³ − 3x² + 2x using the power rule:

dy/dx = 3x² − 6x + 2

Set dy/dx = 0:
3x² − 6x + 2 = 0

Using the quadratic formula:
x = [6 ± √(36 − 24)] / 6 = [6 ± √12] / 6 = [6 ± 2√3] / 6 = (3 ± √3) / 3

Therefore, the x-coordinates where the tangent is horizontal are x = (3 + √3)/3 and x = (3 − √3)/3.

Question 16

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Accepted Answer

This is a related rates problem. Let x = distance of the bottom from the wall, and y = height of the top on the wall.

By the Pythagorean theorem: x² + y² = 100 (since the ladder length is 10 m)

Differentiate both sides with respect to time t:
2x(dx/dt) + 2y(dy/dt) = 0

Simplify:
x(dx/dt) + y(dy/dt) = 0

We are given: dx/dt = 2 m/s

When x = 6 m, find y:
6² + y² = 100
36 + y² = 100
y² = 64
y = 8 m (taking the positive value)

Substitute into the differentiated equation:
6(2) + 8(dy/dt) = 0
12 + 8(dy/dt) = 0
dy/dt = −12/8 = −3/2 = −1.5 m/s

The negative sign indicates the top is sliding down. Therefore, the top of the ladder is sliding down at a rate of 1.5 m/s.

Question 17

The function f(x) = x³ − 6x² + 9x + 1 has a local maximum at x = a and a local minimum at x = b. What is the value of a + b?

Accepted Answer

To find local extrema, we find the critical points by setting f'(x) = 0.

Differentiate f(x) = x³ − 6x² + 9x + 1:
f'(x) = 3x² − 12x + 9

Factor:
f'(x) = 3(x² − 4x + 3) = 3(x − 1)(x − 3)

Critical points: x = 1 and x = 3

To determine which is a maximum and which is a minimum, use the second derivative test:
f''(x) = 6x − 12

At x = 1: f''(1) = 6(1) − 12 = −6 < 0 → local maximum
At x = 3: f''(3) = 6(3) − 12 = 6 > 0 → local minimum

Therefore, a = 1 and b = 3, so a + b = 1 + 3 = 4.

Question 18

A function f(x) = x⁴ − 4x³ + 6x² − 4x + 1 is defined on the interval [0, 2]. What is the absolute maximum value of f on this interval?

Accepted Answer

To find the absolute maximum on a closed interval, we must evaluate f at critical points and endpoints.

First, find critical points by setting f'(x) = 0:
f'(x) = 4x³ − 12x² + 12x − 4
f'(x) = 4(x³ − 3x² + 3x − 1)

Notice that x³ − 3x² + 3x − 1 = (x − 1)³ (by the binomial expansion of (x − 1)³)

So f'(x) = 4(x − 1)³

Setting f'(x) = 0: (x − 1)³ = 0 → x = 1 (with multiplicity 3)

Now evaluate f at the critical point and endpoints:
- At x = 0: f(0) = 0 − 0 + 0 − 0 + 1 = 1
- At x = 1: f(1) = 1 − 4 + 6 − 4 + 1 = 0
- At x = 2: f(2) = 16 − 4(8) + 6(4) − 4(2) + 1 = 16 − 32 + 24 − 8 + 1 = 1

Comparing the values: f(0) = 1, f(1) = 0, f(2) = 1

The absolute maximum value is 1, attained at both x = 0 and x = 2.

Question 19

A particle moves along a curve given by y = x³ − 3x² + 2x. At what point(s) on the curve is the tangent line horizontal?

Accepted Answer

A horizontal tangent occurs when dy/dx = 0. First, differentiate: dy/dx = 3x² − 6x + 2. Setting this equal to zero: 3x² − 6x + 2 = 0. Using the quadratic formula: x = (6 ± √(36 − 24))/6 = (6 ± √12)/6 = (6 ± 2√3)/6 = (3 ± √3)/3. Therefore, the tangent is horizontal at x = (3 + √3)/3 and x = (3 − √3)/3.

Question 20

A cylindrical tank with radius r and height h has volume V = πr²h. If the radius increases at 2 cm/s and the height decreases at 3 cm/s, what is the rate of change of volume (in cm³/s) when r = 5 cm and h = 10 cm?

Accepted Answer

Use the product rule for related rates. Given V = πr²h, differentiate both sides with respect to time t: dV/dt = π[2r(dr/dt)·h + r²(dh/dt)]. Substitute dr/dt = 2, dh/dt = −3, r = 5, h = 10: dV/dt = π[2(5)(2)(10) + (5)²(−3)] = π[200 − 75] = 125π cm³/s.

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