Apply partial differentiation, chain rule, Euler's theorem, and Lagrange multipliers to solve multivariable optimization problems.
A function of two variables \( f(x, y) \) assigns a unique real number \( z = f(x, y) \) to each ordered pair \( (x, y) \) in a domain \( D \subset \mathbb{R}^2 \).
Notation: \( z = f(x, y) \) where \( x \) and \( y \) are independent variables and \( z \) is the dependent variable.
Most engineering quantities depend on multiple variables — temperature \( T(x, y, z, t) \), stress \( \sigma(x, y) \), pressure \( P(V, T) \), etc. Multivariable differentiation studies how these quantities change when one or more variables vary.
We say \( \displaystyle\lim_{(x,y) \to (a,b)} f(x, y) = L \) if for every \( \varepsilon > 0 \), there exists \( \delta > 0 \) such that:
To show a limit does NOT exist, find two different paths that give different limiting values.
Common paths to test:
Problem: Evaluate \( \displaystyle\lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^2 + y^2} \)
Solution:
Using polar coordinates \( x = r\cos\theta \), \( y = r\sin\theta \):
\[ \frac{x^2 y}{x^2 + y^2} = \frac{r^3\cos^2\theta\sin\theta}{r^2} = r\cos^2\theta\sin\theta \]As \( r \to 0 \): \( |r\cos^2\theta\sin\theta| \leq r \to 0 \)
Therefore, the limit exists and equals 0.
Problem: Evaluate \( \displaystyle\lim_{(x,y) \to (0,0)} \frac{xy}{x^2 + y^2} \)
Solution:
Along \( y = mx \):
\[ \frac{xy}{x^2 + y^2} = \frac{mx^2}{x^2 + m^2x^2} = \frac{m}{1 + m^2} \]This depends on \( m \). For \( m = 1 \): value = \( \frac{1}{2} \); for \( m = 2 \): value = \( \frac{2}{5} \)
Different paths give different values → Limit does NOT exist.
A function \( f(x, y) \) is continuous at \( (a, b) \) if:
This requires three conditions:
A partial derivative of \( f(x, y) \) with respect to \( x \) is the derivative of \( f \) treating \( y \) as a constant. Similarly for \( y \).
| Partial w.r.t. \( x \) | Partial w.r.t. \( y \) |
|---|---|
| \( f_x \), \( \dfrac{\partial f}{\partial x} \), \( \dfrac{\partial z}{\partial x} \) | \( f_y \), \( \dfrac{\partial f}{\partial y} \), \( \dfrac{\partial z}{\partial y} \) |
If \( z = f(x, y) \) and \( x, y \) are functions of a parameter \( t \), the total derivative of \( z \) with respect to \( t \) is:
More generally, if \( z = f(x_1, x_2, \ldots, x_n) \) and each \( x_i \) depends on \( t \):
Problem: Find \( f_x \) and \( f_y \) for \( f(x, y) = x^3 y^2 + 2xy + 5 \)
Solution:
With respect to \( x \) (treat \( y \) as constant):
\[ f_x = 3x^2 y^2 + 2y \]With respect to \( y \) (treat \( x \) as constant):
\[ f_y = 2x^3 y + 2x \]Problem: Find all second-order partial derivatives of \( f(x, y) = x^2 y^3 \)
Solution:
First-order: \( f_x = 2xy^3 \), \( f_y = 3x^2 y^2 \)
Second-order:
\[ f_{xx} = \frac{\partial}{\partial x}(2xy^3) = 2y^3 \] \[ f_{yy} = \frac{\partial}{\partial y}(3x^2 y^2) = 6x^2 y \] \[ f_{xy} = \frac{\partial}{\partial y}(2xy^3) = 6xy^2 \] \[ f_{yx} = \frac{\partial}{\partial x}(3x^2 y^2) = 6xy^2 \]Verify: \( f_{xy} = f_{yx} = 6xy^2 \) ✓ (Clairaut's theorem)
Problem: If \( z = x^2 + y^2 \), where \( x = t^2 \) and \( y = 2t \), find \( \dfrac{dz}{dt} \)
Solution:
Using the total derivative formula:
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] \[ \frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y, \quad \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 2 \] \[ \frac{dz}{dt} = (2x)(2t) + (2y)(2) = 4xt + 4y \]Substituting \( x = t^2 \), \( y = 2t \):
\[ \frac{dz}{dt} = 4t^3 + 8t \]Verification: Direct substitution gives \( z = t^4 + 4t^2 \), so \( \dfrac{dz}{dt} = 4t^3 + 8t \) ✓
Problem: If \( x^2 + y^2 + z^2 = 25 \), find \( \dfrac{\partial z}{\partial x} \) and \( \dfrac{\partial z}{\partial y} \)
Solution:
Differentiating with respect to \( x \) (treat \( y \) as constant):
\[ 2x + 2z\frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -\frac{x}{z} \]Similarly:
\[ \frac{\partial z}{\partial y} = -\frac{y}{z} \]The chain rule allows us to differentiate composite functions of several variables. It has several forms depending on the dependency structure.
z
/ \
x y
/ \ / \
s t s t
Each path from \( z \) to \( s \) (or \( t \)) contributes a product of partial derivatives. Sum all paths.
Problem: If \( z = x^2 y + y^3 \), \( x = t^2 \), \( y = t + 1 \), find \( \dfrac{dz}{dt} \)
Solution:
\[ \frac{\partial z}{\partial x} = 2xy, \quad \frac{\partial z}{\partial y} = x^2 + 3y^2 \] \[ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 1 \] \[ \frac{dz}{dt} = (2xy)(2t) + (x^2 + 3y^2)(1) \]Substituting \( x = t^2 \), \( y = t+1 \):
\[ \frac{dz}{dt} = 2(t^2)(t+1)(2t) + t^4 + 3(t+1)^2 = 4t^3(t+1) + t^4 + 3(t+1)^2 \] \[ = 4t^4 + 4t^3 + t^4 + 3t^2 + 6t + 3 = 5t^4 + 4t^3 + 3t^2 + 6t + 3 \]Problem: If \( z = x^2 + y^2 \), \( x = s + t \), \( y = s - t \), find \( \dfrac{\partial z}{\partial s} \) and \( \dfrac{\partial z}{\partial t} \)
Solution:
\[ \frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y \] \[ \frac{\partial x}{\partial s} = 1, \quad \frac{\partial y}{\partial s} = 1, \quad \frac{\partial x}{\partial t} = 1, \quad \frac{\partial y}{\partial t} = -1 \] \[ \frac{\partial z}{\partial s} = (2x)(1) + (2y)(1) = 2x + 2y = 2(s+t) + 2(s-t) = 4s \] \[ \frac{\partial z}{\partial t} = (2x)(1) + (2y)(-1) = 2x - 2y = 2(s+t) - 2(s-t) = 4t \]Problem: If \( w = x^2 + y^2 + z^2 \), \( x = u + v \), \( y = u - v \), \( z = 2uv \), find \( \dfrac{\partial w}{\partial u} \)
Solution:
\[ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u} \] \[ = (2x)(1) + (2y)(1) + (2z)(2v) \] \[ = 2(u+v) + 2(u-v) + 4uv(2v) = 4u + 8uv^2 \]A function \( f(x, y) \) is homogeneous of degree \( n \) if:
where \( n \) is a real number called the degree of homogeneity.
If \( f(x, y) \) is a homogeneous function of degree \( n \) with continuous partial derivatives, then:
For \( f(x, y) \) homogeneous of degree \( n \):
For \( f(x, y, z) \) homogeneous of degree \( n \):
If \( f(x, y) \) is homogeneous of degree \( n \), then its partial derivatives \( f_x \) and \( f_y \) are homogeneous of degree \( n-1 \). Therefore:
Problem: Show that \( f(x, y) = x^3 + 3x^2 y + y^3 \) is homogeneous and find its degree.
Solution:
\[ f(tx, ty) = (tx)^3 + 3(tx)^2(ty) + (ty)^3 = t^3 x^3 + 3t^3 x^2 y + t^3 y^3 \] \[ = t^3(x^3 + 3x^2 y + y^3) = t^3 f(x, y) \]Therefore, \( f \) is homogeneous of degree 3.
Problem: Verify Euler's theorem for \( f(x, y) = x^2 + 3xy + y^2 \).
Solution:
Degree: \( f(tx, ty) = t^2 f(x, y) \) → \( n = 2 \)
Partials: \( f_x = 2x + 3y \), \( f_y = 3x + 2y \)
LHS of Euler's theorem:
\[ x f_x + y f_y = x(2x + 3y) + y(3x + 2y) = 2x^2 + 3xy + 3xy + 2y^2 = 2x^2 + 6xy + 2y^2 \]RHS: \( n f = 2(x^2 + 3xy + y^2) = 2x^2 + 6xy + 2y^2 \)
LHS = RHS ✓ Euler's theorem verified.
Problem: If \( u = \ln\left(\frac{x^3 + y^3}{x + y}\right) \), show that \( x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 2 \).
Solution:
Let \( f = \frac{x^3 + y^3}{x + y} \). Note \( f \) is homogeneous of degree 2.
Then \( u = \ln f \).
\[ \frac{\partial u}{\partial x} = \frac{1}{f}\frac{\partial f}{\partial x}, \quad \frac{\partial u}{\partial y} = \frac{1}{f}\frac{\partial f}{\partial y} \] \[ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{1}{f}\left(x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y}\right) = \frac{1}{f}(2f) = 2 \]Verified ✓
Problem: If \( u = x^2 + y^2 + z^2 + 2xyz \), find \( x u_x + y u_y + z u_z \).
Solution:
Check homogeneity: \( u(tx, ty, tz) = t^2(x^2 + y^2 + z^2 + 2xyz) = t^2 u \) → degree \( n = 2 \)
By Euler's theorem:
\[ x u_x + y u_y + z u_z = 2u = 2(x^2 + y^2 + z^2 + 2xyz) \]A function \( f(x, y) \) has a local maximum at \( (a, b) \) if \( f(a, b) \geq f(x, y) \) for all \( (x, y) \) near \( (a, b) \). Similarly for a local minimum.
A point where either \( f_x = f_y = 0 \) or one of the partials doesn't exist is called a critical point.
Let \( (a, b) \) be a critical point where \( f_x(a,b) = f_y(a,b) = 0 \).
Define:
| Condition | Conclusion |
|---|---|
| \( D > 0 \) and \( f_{xx}(a,b) > 0 \) | Local Minimum |
| \( D > 0 \) and \( f_{xx}(a,b) < 0 \) | Local Maximum |
| \( D < 0 \) | Saddle Point (neither max nor min) |
| \( D = 0 \) | Inconclusive — further investigation needed |
Problem: Find the extreme values of \( f(x, y) = x^2 + y^2 - 2x - 4y + 5 \)
Solution:
Step 1: \( f_x = 2x - 2 \), \( f_y = 2y - 4 \)
Step 2: \( f_x = 0 \Rightarrow x = 1 \); \( f_y = 0 \Rightarrow y = 2 \)
Critical point: \( (1, 2) \)
Step 3: \( f_{xx} = 2 \), \( f_{yy} = 2 \), \( f_{xy} = 0 \)
Step 4: \( D = (2)(2) - 0^2 = 4 > 0 \), \( f_{xx} = 2 > 0 \)
Step 5: Local minimum at \( (1, 2) \)
Step 6: \( f(1, 2) = 1 + 4 - 2 - 8 + 5 = 0 \)
Minimum value = 0 at \( (1, 2) \)
Problem: Find the extreme values of \( f(x, y) = 4x - x^2 - y^2 + 2y - 1 \)
Solution:
\( f_x = 4 - 2x \), \( f_y = -2y + 2 \)
\( f_x = 0 \Rightarrow x = 2 \); \( f_y = 0 \Rightarrow y = 1 \)
Critical point: \( (2, 1) \)
\( f_{xx} = -2 \), \( f_{yy} = -2 \), \( f_{xy} = 0 \)
\( D = (-2)(-2) - 0 = 4 > 0 \), \( f_{xx} = -2 < 0 \)
Local maximum at \( (2, 1) \)
\( f(2, 1) = 8 - 4 - 1 + 2 - 1 = 4 \)
Maximum value = 4 at \( (2, 1) \)
Problem: Find the extreme values of \( f(x, y) = x^2 - y^2 \)
Solution:
\( f_x = 2x \), \( f_y = -2y \)
Critical point: \( (0, 0) \)
\( f_{xx} = 2 \), \( f_{yy} = -2 \), \( f_{xy} = 0 \)
\( D = (2)(-2) - 0 = -4 < 0 \)
Saddle point at \( (0, 0) \) — no extremum
Problem: Find the extreme values of \( f(x, y) = x^3 + y^3 - 3xy \)
Solution:
Step 1: \( f_x = 3x^2 - 3y \), \( f_y = 3y^2 - 3x \)
Step 2: Setting both to zero: \( y = x^2 \) and \( x = y^2 \)
Substituting: \( x = (x^2)^2 = x^4 \Rightarrow x^4 - x = 0 \Rightarrow x(x^3 - 1) = 0 \)
\( x = 0 \) or \( x = 1 \)
Critical points: \( (0, 0) \) and \( (1, 1) \)
Step 3: \( f_{xx} = 6x \), \( f_{yy} = 6y \), \( f_{xy} = -3 \)
At \( (0, 0) \): \( D = (0)(0) - (-3)^2 = -9 < 0 \) → Saddle point
At \( (1, 1) \): \( D = (6)(6) - (-3)^2 = 36 - 9 = 27 > 0 \), \( f_{xx} = 6 > 0 \) → Local minimum
\( f(1, 1) = 1 + 1 - 3 = -1 \)
Local minimum value = -1 at \( (1, 1) \)
Often we need to optimize a function \( f(x, y) \) subject to a constraint \( g(x, y) = 0 \). The Lagrange multiplier method handles such problems elegantly.
To optimize \( f(x, y) \) subject to \( g(x, y) = 0 \):
To optimize \( f(x, y, z) \) subject to \( g(x, y, z) = 0 \):
Set \( \mathcal{L}_x = \mathcal{L}_y = \mathcal{L}_z = \mathcal{L}_\lambda = 0 \).
Problem: Find the maximum value of \( f(x, y) = xy \) subject to \( x + y = 10 \).
Solution:
Step 1: Constraint: \( g(x, y) = x + y - 10 = 0 \)
Step 2: Lagrangian:
\[ \mathcal{L} = xy + \lambda(x + y - 10) \]Step 3: Partial derivatives:
\[ \frac{\partial \mathcal{L}}{\partial x} = y + \lambda = 0 \implies y = -\lambda \] \[ \frac{\partial \mathcal{L}}{\partial y} = x + \lambda = 0 \implies x = -\lambda \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = x + y - 10 = 0 \]Step 4: From first two equations: \( x = y \). Substituting into constraint:
\[ 2x = 10 \implies x = 5, \; y = 5 \]Step 5: \( f(5, 5) = 25 \)
Maximum value = 25 at \( (5, 5) \)
Problem: Find the maximum value of \( f(x, y, z) = xyz \) subject to \( x + y + z = 9 \).
Solution:
Lagrangian: \( \mathcal{L} = xyz + \lambda(x + y + z - 9) \)
Partials:
\[ yz + \lambda = 0, \quad xz + \lambda = 0, \quad xy + \lambda = 0 \]From first two: \( yz = xz \Rightarrow x = y \) (since \( z \neq 0 \))
From last two: \( xz = xy \Rightarrow z = y \)
Therefore \( x = y = z \). Constraint: \( 3x = 9 \Rightarrow x = 3 \)
Maximum value = \( 3 \times 3 \times 3 = 27 \) at \( (3, 3, 3) \)
Problem: Find the dimensions of a rectangular box of maximum volume with total surface area 150 cm².
Solution:
Let dimensions be \( x, y, z \). Maximize \( V = xyz \) subject to \( 2(xy + yz + zx) = 150 \), i.e., \( xy + yz + zx = 75 \).
Lagrangian: \( \mathcal{L} = xyz + \lambda(xy + yz + zx - 75) \)
Partials:
\[ yz + \lambda(y + z) = 0 \quad \text{(1)} \] \[ xz + \lambda(x + z) = 0 \quad \text{(2)} \] \[ xy + \lambda(x + y) = 0 \quad \text{(3)} \]From (1) and (2): \( yz - xz = \lambda(x - y) \Rightarrow z(y - x) = \lambda(x - y) \Rightarrow \lambda = -z \)
Similarly from (2) and (3): \( \lambda = -x \). From (1) and (3): \( \lambda = -y \).
Therefore \( x = y = z \). Constraint: \( 3x^2 = 75 \Rightarrow x^2 = 25 \Rightarrow x = 5 \)
Box dimensions: 5 cm × 5 cm × 5 cm (a cube)
Maximum volume = 125 cm³
| Topic | Key Formula / Concept |
|---|---|
| Limit of \( f(x,y) \) | Same value along every path approaching \( (a,b) \) |
| Continuity | \( \lim_{(x,y)\to(a,b)} f(x,y) = f(a,b) \) |
| Partial Derivative \( f_x \) | Differentiate w.r.t. \( x \), treat \( y \) as constant |
| Partial Derivative \( f_y \) | Differentiate w.r.t. \( y \), treat \( x \) as constant |
| Clairaut's Theorem | \( f_{xy} = f_{yx} \) (for continuous mixed partials) |
| Total Derivative | \( \dfrac{dz}{dt} = \dfrac{\partial z}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial z}{\partial y}\dfrac{dy}{dt} \) |
| Chain Rule (2 params) | \( \dfrac{\partial z}{\partial s} = \dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial s} + \dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial s} \) |
| Homogeneous Function | \( f(tx, ty) = t^n f(x, y) \) |
| Euler's Theorem | \( x f_x + y f_y = n f \) |
| Euler's Theorem (3 var) | \( x f_x + y f_y + z f_z = n f \) |
| Critical Point | \( f_x = f_y = 0 \) |
| Second Derivative Test | \( D = f_{xx} f_{yy} - f_{xy}^2 \) |
| Local Minimum | \( D > 0 \), \( f_{xx} > 0 \) |
| Local Maximum | \( D > 0 \), \( f_{xx} < 0 \) |
| Saddle Point | \( D < 0 \) |
| Inconclusive | \( D = 0 \) |
| Lagrangian | \( \mathcal{L} = f + \lambda g \) |
| Lagrange Conditions | \( \mathcal{L}_x = \mathcal{L}_y = \mathcal{L}_\lambda = 0 \) |
| Function \( f(x, y) \) | \( f_x \) | \( f_y \) |
|---|---|---|
| \( x^n y^m \) | \( nx^{n-1}y^m \) | \( mx^n y^{m-1} \) |
| \( e^{ax + by} \) | \( a e^{ax+by} \) | \( b e^{ax+by} \) |
| \( \ln(x^2 + y^2) \) | \( \dfrac{2x}{x^2+y^2} \) | \( \dfrac{2y}{x^2+y^2} \) |
| \( \sin(xy) \) | \( y\cos(xy) \) | \( x\cos(xy) \) |
| \( \tan^{-1}(y/x) \) | \( -\dfrac{y}{x^2+y^2} \) | \( \dfrac{x}{x^2+y^2} \) |
| \( x^y \) | \( yx^{y-1} \) | \( x^y \ln x \) |
Evaluate \( \displaystyle\lim_{(x,y) \to (0,0)} \frac{x^2 - y^2}{x^2 + y^2} \) or show it does not exist.
Find \( f_x \) and \( f_y \) for \( f(x, y) = x^2 e^{xy} + y^3 \ln x \).
If \( z = x^2 y + y^2 \), \( x = t^2 + 1 \), \( y = 2t - 1 \), find \( \dfrac{dz}{dt} \).
If \( u = x^3 + y^3 + z^3 - 3xyz \), find \( x u_x + y u_y + z u_z \).
Find the extreme values of \( f(x, y) = x^2 + y^2 - 4x - 6y + 10 \).
Find the critical points of \( f(x, y) = x^3 + y^3 - 6xy \) and classify them.
Find the maximum value of \( f(x, y) = x + 2y \) subject to \( x^2 + y^2 = 5 \).
Find the dimensions of a rectangular box with maximum volume and total surface area 96 cm².
Along \( y = 0 \): \( \frac{x^2}{x^2} = 1 \)
Along \( x = 0 \): \( \frac{-y^2}{y^2} = -1 \)
Different paths give different values → Limit does not exist.
\( f_x = 2x e^{xy} + x^2 y e^{xy} + \dfrac{y^3}{x} = e^{xy}(2x + x^2 y) + \dfrac{y^3}{x} \)
\( f_y = x^3 e^{xy} + 3y^2 \ln x \)
\( \dfrac{\partial z}{\partial x} = 2xy \), \( \dfrac{\partial z}{\partial y} = x^2 + 2y \)
\( \dfrac{dx}{dt} = 2t \), \( \dfrac{dy}{dt} = 2 \)
\[ \frac{dz}{dt} = (2xy)(2t) + (x^2 + 2y)(2) = 4xyt + 2x^2 + 4y \]Substituting \( x = t^2+1, y = 2t-1 \):
\[ = 4(t^2+1)(2t-1)t + 2(t^2+1)^2 + 4(2t-1) \]Check homogeneity: \( u(tx, ty, tz) = t^3 u(x,y,z) \) → \( n = 3 \)
By Euler's theorem:
\[ x u_x + y u_y + z u_z = 3u = 3(x^3 + y^3 + z^3 - 3xyz) \]\( f_x = 2x - 4 = 0 \Rightarrow x = 2 \)
\( f_y = 2y - 6 = 0 \Rightarrow y = 3 \)
Critical point: \( (2, 3) \)
\( f_{xx} = 2 \), \( f_{yy} = 2 \), \( f_{xy} = 0 \)
\( D = 4 > 0 \), \( f_{xx} = 2 > 0 \) → Local minimum
\( f(2,3) = 4 + 9 - 8 - 18 + 10 = -3 \)
Minimum value = -3 at (2, 3)
\( f_x = 3x^2 - 6y = 0 \Rightarrow y = \frac{x^2}{2} \)
\( f_y = 3y^2 - 6x = 0 \Rightarrow x = \frac{y^2}{2} \)
Substituting: \( x = \frac{(x^2/2)^2}{2} = \frac{x^4}{8} \Rightarrow x^4 = 8x \Rightarrow x(x^3 - 8) = 0 \)
\( x = 0 \) or \( x = 2 \)
Critical points: \( (0, 0) \) and \( (2, 2) \)
\( f_{xx} = 6x \), \( f_{yy} = 6y \), \( f_{xy} = -6 \)
At (0,0): \( D = 0 - 36 = -36 < 0 \) → Saddle point
At (2,2): \( D = (12)(12) - 36 = 108 > 0 \), \( f_{xx} = 12 > 0 \) → Local minimum
\( f(2,2) = 8 + 8 - 24 = -8 \) → Minimum value = -8
\( \mathcal{L} = x + 2y + \lambda(x^2 + y^2 - 5) \)
\( 1 + 2\lambda x = 0 \Rightarrow x = -\dfrac{1}{2\lambda} \)
\( 2 + 2\lambda y = 0 \Rightarrow y = -\dfrac{1}{\lambda} \)
Substituting into \( x^2 + y^2 = 5 \): \( \dfrac{1}{4\lambda^2} + \dfrac{1}{\lambda^2} = 5 \Rightarrow \dfrac{5}{4\lambda^2} = 5 \Rightarrow \lambda^2 = \dfrac{1}{4} \)
\( \lambda = \pm\dfrac{1}{2} \). For maximum, take \( \lambda = -\dfrac{1}{2} \): \( x = 1, y = 2 \)
\( f(1, 2) = 1 + 4 = 5 \) → Maximum = 5
Maximize \( V = xyz \) subject to \( 2(xy + yz + zx) = 96 \Rightarrow xy + yz + zx = 48 \)
By symmetry, \( x = y = z \). Then \( 3x^2 = 48 \Rightarrow x^2 = 16 \Rightarrow x = 4 \)
Dimensions: 4 cm × 4 cm × 4 cm (cube)
Maximum volume = \( 4^3 = 64 \) cm³
| Reference | Book |
|---|---|
| T-1 | Advanced Engineering Mathematics – R.K. Jain, S.R.K. Iyengar (Narosa Publishing House) |
| R-1 | Higher Engineering Mathematics – B.S. Grewal (Khanna Publishers) |
| R-2 | Mathematics Textbook for Class XII Part I – NCERT |
| R-3 | Mathematics Textbook for Class XII Part II – NCERT |
| OR-1 | https://ncert.nic.in/textbook.php?lemh1=5-6 |
| OR-2 | https://ncert.nic.in/textbook.php?lemh2=1-7 |
After studying this unit, students should be able to:
| CO4 Components | Covered in Section |
|---|---|
| Apply partial differentiation | III |
| Apply chain rule | IV |
| Apply Euler's theorem | V |
| Solve maxima/minima problems | VI |
| Apply Lagrange multipliers | VII |
| Understand limits and continuity | II |
⭐ End of Unit IV Notes ⭐
Multivariable Differentiation
MTH165 – Mathematics for Engineers
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