Graded Assignment 11

Question 1

Problem: Find the correct second partial derivative $f_{xx}$ for $f(x, y, z) = x^2z + y^3e^{xy} + \sin(xyz)$.

Concept: Partial Differentiation To find $f_{xx}$, we first differentiate $f$ with respect to $x$ (treating $y, z$ as constants), and then differentiate the result by $x$ again.

Step 1: First Derivative ($f_x$)

• $\frac{\partial}{\partial x}(x^2z) = 2xz$
• $\frac{\partial}{\partial x}(y^3e^{xy}) = y^3 \cdot \frac{\partial}{\partial x}(e^{xy}) = y^3 \cdot (ye^{xy}) = y^4e^{xy}$
• $\frac{\partial}{\partial x}(\sin(xyz)) = \cos(xyz) \cdot \frac{\partial}{\partial x}(xyz) = yz\cos(xyz)$

Step 2: Second Derivative ($f_{xx}$)

• $\frac{\partial}{\partial x}(2xz) = 2z$
• $\frac{\partial}{\partial x}(y^4e^{xy}) = y^4 \cdot (ye^{xy}) = y^5e^{xy}$
• $\frac{\partial}{\partial x}(yz\cos(xyz)) = yz \cdot (-\sin(xyz) \cdot yz) = -y^2z^2\sin(xyz)$

Answer Option: $f_{xx} = 2z + y^5e^{xy} - (yz)^2 \sin(xyz)$ (Second Option)

Question 2

Problem: Number of critical points of $f(x, y, z) = e^x(x^2 - y^2 - 2z^2)$.

Concept: Critical Points Critical points occur where the gradient is zero ($\nabla f = \vec{0}$). We set partial derivatives $f_x, f_y, f_z$ to 0.

Step 1: Partial Derivatives

• $f_y = e^x(-2y)$. Setting $f_y = 0 \implies y=0$.
• $f_z = e^x(-4z)$. Setting $f_z = 0 \implies z=0$.
• $f_x = \frac{\partial}{\partial x}(e^x) \cdot (x^2 - y^2 - 2z^2) + e^x \cdot \frac{\partial}{\partial x}(x^2 - y^2 - 2z^2)$ $f_x = e^x(x^2 - y^2 - 2z^2) + e^x(2x) = e^x(x^2 + 2x - y^2 - 2z^2)$.

Step 2: Solve System Substitute $y=0, z=0$ into $f_x=0$:

Since $e^x \neq 0$, we solve $x^2 + 2x = 0$:

Step 3: Count Points The critical points are $(0, 0, 0)$ and $(-2, 0, 0)$.

Answer: 2

Question 3

Problem: Analyze $f(x, y) = \sqrt{x^2 + y^2}$ at $(0,0)$.

Concept: Geometry of the Cone This function represents an inverted cone with the tip at the origin.

• Global Minimum: Since $\sqrt{\dots} \ge 0$ and $f(0,0)=0$, the point $(0,0)$ is the absolute lowest point.
• Differentiability: The cone has a sharp “cusp” or point at the origin. It is continuous but not differentiable at $(0,0)$. (Imagine trying to place a tangent plane on the tip of a sharp pencil—it wobbles).

Correct Statements:

• $(0,0)$ is a local minimum of $f$.
• $(0,0)$ is a global minimum of $f$.
• $(0,0)$ is the only critical point (where derivative is undefined).

Answer Selection: $(0,0)$ is a global minimum of $f$.

Question 4

Problem: Find the minimum surface area of a rectangular box with Volume = 1000.

Concept: Optimization For a rectangular box with fixed volume, the surface area is minimized when the box is a cube. Let sides be $x, y, z$. Volume $V = xyz = 1000$. Surface Area $S = 2(xy + yz + zx)$.

Calculation: For a cube, $x = y = z$.

Answer: 600

Question 5

Problem: Hessian Matrix properties for $f(x,y,z) = x^3 + 3y^3 + z^3 - 4x - 6y - 4z$ at $(0,1,1)$.

Concept: Hessian Matrix The Hessian is the matrix of second-order partial derivatives.

Step 1: First Derivatives $f_x = 3x^2 - 4$ $f_y = 9y^2 - 6$ $f_z = 3z^2 - 4$

Step 2: Second Derivatives $f_{xx} = 6x$, $f_{yy} = 18y$, $f_{zz} = 6z$. All mixed terms ($f_{xy}$, etc.) are 0.

Step 3: Evaluate at $(0,1,1)$ $f_{xx}(0) = 0$ $f_{yy}(1) = 18$ $f_{zz}(1) = 6$

Step 4: Matrix Properties

• Determinant: Product of diagonal = $0 \times 18 \times 6 = 0$.
• Rank: Number of non-zero rows = 2.
• Nullity: Dimension - Rank = $3 - 2 = 1$.

Correct Options:

• $\det(A) = 0$
• $Rank(A) = 2$
• $Nullity(A) = 1$

Question 8

Problem: Max value of $-2x^2 + 2xy - y^2 + 12x - 4y - 7$.

Concept: Unconstrained Optimization Find critical points ($f_x=0, f_y=0$).

Step 1: Partials $f_x = -4x + 2y + 12 = 0 \implies 2x - y = 6$ (Eq 1) $f_y = 2x - 2y - 4 = 0 \implies x - y = 2$ (Eq 2)

Step 2: Solve Subtract Eq 2 from Eq 1: $(2x - y) - (x - y) = 6 - 2$ $x = 4$ Sub into Eq 2: $4 - y = 2 \implies y = 2$

Step 3: Evaluate $f(4,2) = -2(16) + 2(4)(2) - (4) + 12(4) - 4(2) - 7$ $= -32 + 16 - 4 + 48 - 8 - 7$ $= 13$

Answer: 13

Question 10

Problem: Maximize Profit. Revenue $R = 2x + 3y$. Cost $C = 2x^2 - 2xy + y^2 - 9x + 6y + 7$. Profit $P = R - C$.

Step 1: Profit Function $P(x,y) = (2x + 3y) - (2x^2 - 2xy + y^2 - 9x + 6y + 7)$ $P = -2x^2 + 2xy - y^2 + 11x - 3y - 7$

Step 2: Gradients $P_x = -4x + 2y + 11 = 0$ $P_y = 2x - 2y - 3 = 0$

Step 3: Solve Add the two equations: $(-4x + 2y + 11) + (2x - 2y - 3) = 0$ $-2x + 8 = 0 \implies x = 4$. Sub $x=4$ into $P_y$: $2(4) - 2y - 3 = 0 \implies 5 = 2y \implies y = 2.5$.

Step 4: Max Profit $P(4, 2.5) = -2(16) + 2(4)(2.5) - (2.5)^2 + 11(4) - 3(2.5) - 7$ $= -32 + 20 - 6.25 + 44 - 7.5 - 7$ $= 11.25$

Answer: 11.25

Question 12

Problem: Find stable equilibrium point (Potential Energy at Minimum) for $V(x,y) = x^4 + y^4 - 4xy + 1$.

Step 1: Critical Points $V_x = 4x^3 - 4y = 0 \implies y = x^3$ $V_y = 4y^3 - 4x = 0 \implies x = y^3$ Sub $y$: $x = (x^3)^3 = x^9 \implies x(x^8-1) = 0$. $x = 0, 1, -1$. Points: $(0,0), (1,1), (-1,-1)$.

Step 2: Hessian Test $V_{xx} = 12x^2, V_{yy} = 12y^2, V_{xy} = -4$. Determinant $D = 144x^2y^2 - 16$.

• At (0,0): $D = -16 < 0$ (Saddle).
• At (1,1): $D = 144-16 > 0$ and $V_{xx}=12 > 0$ (Local Min).
• At (-1,-1): $D = 144-16 > 0$ and $V_{xx}=12 > 0$ (Local Min).

Step 3: Potential Energy Value $V(1,1) = 1 + 1 - 4 + 1 = -1$. $V(-1,-1) = 1 + 1 - 4 + 1 = -1$.

Answer: -1.00

Case Study: Fitness App (Q9-Q19, Q20-Q27)

Context: $\mu=100, \sigma=2, n=9$. Acceptance Region: $98.5 \le \bar{X} \le 101.5$. Reject if $\bar{X} < 98.5$ or $\bar{X} > 101.5$.

Q10-Q19: Significance Level ($\alpha$)

1. Q10 (Blank 1): Sampling dist is Normal (Option 2).
2. Q11 (Blank 2): Mean is 100 (Option 4).
3. Q12 (Blank 3): Variance is $\sigma^2/n = 4/9$ (Option 6).
4. Q13 (Blank 4): Reject if $\bar{X} >$ 101.5 (Option 8).
5. Q14 (Blank 5): Reject if $\bar{X} <$ 98.5 (Option 7).
6. Q16 (Blank 7): Critical Z = $\frac{101.5 - 100}{\sqrt{4/9}} = \frac{1.5}{2/3} = 2.25$. 2.25 (Option 13).
7. Q17 (Blank 8): Lower Z = -2.25 (Option 12).
8. Q19 (Blank 10): $\alpha = 1 - P(-2.25 < Z < 2.25) \approx 1 - 0.975 = 0.025$. Answer 0.02 (Option 14).

Q20-Q27: Power ($1-\beta$) at $\mu=103$

1. Q21 (Blank 1): Upper boundary 101.5 (Option 3).
2. Q22 (Blank 2): Lower boundary 98.5 (Option 1).
3. Q23 (Blank 3): Standardized Upper $Z = \frac{101.5 - 103}{2/3} = -2.25$. -2.25 (Option 12).
4. Q24 (Blank 4): Standardized Lower $Z = \frac{98.5 - 103}{2/3} = -6.75$. -6.75 (Option 7).
5. Q25 (Blank 5): $P(Z > -2.25) = P(Z < 2.25) \approx$ 0.99 (Option 10).
6. Q27 (Blank 7): Power = 0.99 + 0.

Question 10 (Independent)

Problem: Sample Size for Z-test. $X \sim N(\mu_1, 3)$ and $Y \sim N(\mu_2, 4)$. (Note: 3 and 4 are Variances). Rejection rule: $|\bar{X} - \bar{Y}| > 1.625$. $\alpha=0.05 \implies Z_{crit}=1.96$.

Formula: Critical Value $c = Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}$ $1.625 = 1.96 \sqrt{\frac{3}{n} + \frac{4}{8}}$

Calculation: Divide by 1.96: $0.829 = \sqrt{\frac{3}{n} + 0.5}$ Square both sides: $0.687 = \frac{3}{n} + 0.5$ $\frac{3}{n} = 0.187$ $n = \frac{3}{0.187} \approx 16.04$

Answer: 17 (Next greatest integer)

Question 11 (Battery Case Study)

Problem: Find critical value $c$ for t-test. $H_0: \mu = 11.5$, $H_A: \mu > 11.5$. $n=25, s=2.5$. $\alpha=0.05$. Degrees of freedom = 24. Critical $t$ from table ($F_{t24}(1.711) = 0.95$) is 1.711.

Calculation: $\frac{c - 11.5}{s/\sqrt{n}} = 1.711$ $c - 11.5 = 1.711 \times \frac{2.5}{5}$ $c - 11.5 = 1.711 \times 0.5$ $c = 11.5 + 0.8555 = 12.3555$

Answer: 12.36

Question 7 (Drug Variance)

Problem: Identify Hypothesis. Pharmacist suspects variability ($\sigma$) is higher than expected ($0.00002$).

Concept:

• Null Hypothesis ($H_0$): Status quo ($\sigma = 0.00002$).
• Alternative ($H_A$): The claim/suspicion ($\sigma > 0.00002$).

Answer: $H_0: \sigma = 0.00002, H_A: \sigma > 0.00002$ (Option 1)

Question 8 (Drug Variance Test)

Problem: Chi-Square test for Q7. $n=8, s=0.00005, \sigma_0=0.00002$.

Calculation: $\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{7 \cdot (5 \times 10^{-5})^2}{(2 \times 10^{-5})^2} = 7 \cdot (2.5)^2 = 7 \cdot 6.25 = 43.75$. Critical Value (given hint): 18.475. Since $43.75 > 18.475$, we Reject $H_0$ and Accept $H_A$.

Answer: Accept the alternative hypothesis.