Question 1

Find the critical points of f(x, y) = 3x² - 2xy + 2y² - 10x + 1. A) (2, 1) B) (1, 1) C) (0, 0) D) (1, -1) and (-1, 1) E) (2, 1) and (1, 2)

Accepted Answer

To find the critical points of $f(x, y) = 3x^2 - 2xy + 2y^2 - 10x + 1$, we first find the partial derivatives of $f$ with respect to $x$ and $y$, set them equal to zero, and solve the system of equations. The partial derivative with respect to $x$ is $f_x = 6x - 2y - 10$, and the partial derivative with respect to $y$ is $f_y = -2x + 4y$. Setting these equal to zero gives the system of equations:$$6x - 2y - 10 = 0$$$$-2x + 4y = 0$$Solving this system yields $x = 2$ and $y = 1$, which corresponds to choice (A).

Question 2

Find the critical points of f(x, y) = ln (x² + y² + 4x - 4y + 8). A) (-2, 2) B) (2, -2) C) (-2, 2) and (2, -2) D) (-2, 2), (2, -2), and (0, 0) E) There are no critical points.

Accepted Answer

E

Question 3

Find all critical points of f(x) = 2x³y -4x³ + 6y³ -18y + 19. A) (0, 1), (0, - 1), and (-9, 2) B) only (-3, 2) C) only (3, 2) D) (0, 1), (0, -1), and (-3, 2) E) only (0, 1) and (-3, 2)

Accepted Answer

To find the critical points of the function $f(x, y) = 2x^3 - 4x + 6y^3 - 18y + 19$, we need to find the points where the gradient of $f$ (denoted as $\nabla f$) is zero. The gradient of $f$ is given by the partial derivatives of $f$ with respect to $x$ and $y$, respectively. 1. Calculate $\frac{\partial f}{\partial x} = 6x^2 - 4$ and set it equal to 0 to solve for $x$.2. Calculate $\frac{\partial f}{\partial y} = 18y^2 - 18$ and set it equal to 0 to solve for $y$.Solving $\frac{\partial f}{\partial x} = 0$ gives $x = \pm \sqrt{\frac{2}{3}}$, but this does not match any of the options provided, indicating a mistake in interpreting the superscript notation. Correctly interpreting the notation as $2x^3 - 4x + 6y^3 - 18y + 19$, and solving the partial derivatives:1. $\frac{\partial f}{\partial x} = 6x^2 - 4 = 0$ leads to $x = \pm \sqrt{\frac{2}{3}}$, which was incorrectly considered before. Correctly solving it should give $x = \pm 1$, but this is also incorrect based on the options. The correct approach is to solve $6x^2 - 4 = 0$ without misinterpretation, which actually does not directly lead to the options provided, indicating a mistake in calculation or interpretation in this explanation step.2. $\frac{\partial f}{\partial y} = 18y^2 - 18 = 0$ simplifies to $y^2 = 1$, giving $y = \pm 1$.Given the error in the explanation for $x$, the correct approach is to re-evaluate the partial derivatives and their solutions:- For $x$, solving $6x^2 - 4 = 0$ correctly should give specific values that match the options.- For $y$, correctly solving $18y^2 - 18 = 0$ indeed gives $y = \pm 1$.However, the correct critical points must be determined by accurately solving both equations and checking against the options provided. The mistake in the explanation regarding the solution for $x$ suggests a misunderstanding or misinterpretation of the original function or the process of finding critical points.Given the options and the typical process for finding critical points, without the correct solutions for $x$ from the partial derivative with respect to $x$, it's challenging to directly validate the options. However, the correct approach involves:1. Correctly solving the partial derivatives for $x$ and $y$.2. Identifying the points where both partial derivatives are zero.The presence of a mistake in interpreting or solving the partial derivatives indicates the explanation provided does not accurately reflect the correct method for finding critical points of the given function. The correct critical points would be determined by accurately solving the equations derived from setting the partial derivatives to zero and matching those solutions to the provided options.

Question 4

Find and classify the critical points of the function f(x, y) = x sin y.
A) local maxima at (0, n $\pi$) for even integers n, local minimum at (0, n $\pi$) for odd integers n
B) local minima at (0, n $\pi$) for even integers n, local maximum at (0, n $\pi$) for odd integers n
C) saddle points at all points (0, n $\pi$) for integers n
D) only critical point is (0, 0), a saddle point
E) local minima at (0, n $\pi$) for nonzero integers n, saddle point at (0, 0)

Accepted Answer

To find the critical points of $f(x, y) = x \sin y$, we first find the partial derivatives: $f_x = \sin y$ and $f_y = x \cos y$. Setting these equal to zero, we find that $f_x = 0$ for all $y$, and $f_y = 0$ when $x = 0$ or $\cos y = 0$. However, $f_x = 0$ does not depend on $x$, so we focus on $f_y = 0$, which gives critical points at $x = 0$ and $y = n\pi$ for integers $n$. To classify these points, we examine the second derivatives and the Hessian matrix. The second derivatives are $f_{xx} = 0$, $f_{yy} = -x \sin y$, and $f_{xy} = \cos y$. At $x = 0$, $f_{yy} = 0$, making the Hessian determinant $0$, which does not conclusively classify these points. However, considering the nature of the function and its symmetry, we can infer that these points do not lead to local maxima or minima but rather to changes in concavity, suggesting they are saddle points. Thus, the correct classification is that all points $(0, n\pi)$ for integers $n$ are saddle points.

Question 5

Find and classify the critical points of the function f(x, y) = 2y³ - 3x² - 3xy + 9x. A) local minimum at (2, -1), saddle point at B) local maximum at (2, -1), saddle point at C) saddle point at (2, -1), local maximum at D) saddle point at (2, -1), local minimum at E) saddle point at (2, -1) and

Accepted Answer

The answer of Find and classify the critical points of...

Question 6

Find and classify all critical points for the function f(x, y) =  .
A) (1, 0); minimum
B) (-1, 0); maximum
C) (0, 0); minimum
D) (0, 0); saddle
E) There are no critical points.

Accepted Answer

The answer of Find and classify all critical points for...

Question 7

Find and classify all critical points for the function f(x, y) = x³ - 12xy² + y³ + 45y. A) (2, 1) and (-2, -1) are saddle points B) (2, 1), (2, -1), (-2, 1), and (-2, -1) are saddle points C) (2, 1) is a local maximum and (-2, -1) is a local minimum D) (2, -1) and (-2, 1) are saddle points E) There are no critical points.

Accepted Answer

The critical points are found by solving the system of equations $$f_x = 3x^2 - 12y = 0$$ $$f_y = -24xy + 3y^2 + 45 = 0$$ Solving this system gives the critical points (2, 1) and (-2, -1). To classify these critical points, we compute the discriminant $$D = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(-6y) - (-12y)^2 = -36xy + 144y^2$$ At (2, 1), D = -36(2)(1) + 144(1)^2 = 108 > 0 and $$f_{xx} = 6x = 6(2) = 12 > 0$$ so (2, 1) is a local maximum. At (-2, -1), D = -36(-2)(-1) + 144(-1)^2 = 108 > 0 and $$f_{xx} = 6x = 6(-2) = -12 < 0$$ so (-2, -1) is a local minimum.

Question 8

Find the maximum and minimum values of f(x, y) = (x - y)  .
A) maximum  , minimum 0
B) maximum  , minimum 
C) maximum   , minimum -  
D) maximum 2  , minimum 0
E) maximum  , minimum -

Accepted Answer

The answer of Find the maximum and minimum values of...

Question 9

Find and classify all critical points for the function f(x, y) = 2y³ + 3y² - 12y -x² + 2x. A) (1, -2) local maximum; (1, 1) local minimum B) (1, -2) local maximum; (1, 1) saddle point C) (1, -2) local maximum; (1, 1) local maximum D) (1, -2) local minimum; (1, 1) local minimum E) (-1, 2) saddle point; (1, 1) saddle point

Accepted Answer

To find and classify the critical points of the function $f(x, y) = 2y^3 + 3y^2 - 12y - x^2 + 2x$, we first find the partial derivatives and set them equal to zero to find critical points. Then, we use the second derivative test to classify these points.1. Find the partial derivatives:$\frac{\partial f}{\partial x} = -2x + 2$,$\frac{\partial f}{\partial y} = 6y^2 + 6y - 12$.2. Set the partial derivatives equal to zero to find critical points:$-2x + 2 = 0$ gives $x = 1$,$6y^2 + 6y - 12 = 0$ simplifies to $y^2 + y - 2 = 0$, which factors to $(y + 2)(y - 1) = 0$, giving $y = -2$ and $y = 1$.So, the critical points are $(1, -2)$ and $(1, 1)$.3. To classify these points, we use the second derivative test, which involves computing the second partial derivatives and evaluating the determinant of the Hessian matrix at each critical point.The second partial derivatives are:$\frac{\partial^2 f}{\partial x^2} = -2$,$\frac{\partial^2 f}{\partial y^2} = 12y + 6$,$\frac{\partial^2 f}{\partial x \partial y} = 0$.The Hessian determinant at a point $(x, y)$ is given by $D = f_{xx}f_{yy} - (f_{xy})^2$.For $(1, -2)$, $D = (-2)(12(-2) + 6) - 0^2 = -2(-18) = 36 > 0$ and $f_{xx} = -2 < 0$, indicating a local maximum.For $(1, 1)$, $D = (-2)(12(1) + 6) - 0^2 = -2(18) = -36 < 0$, indicating a saddle point.Therefore, the correct answer is (1, -2) is a local maximum and (1, 1) is a saddle point.

Question 10

A rectangular box with no top has given surface area S. Find the length, width, and height of the box if the volume is as large as possible.
A) length = width =  , height = 
B) length = width =  , height = 
C) length = width =  , height = 
D) length = width =  , height = 
E) length = width =  , height =

Accepted Answer

The answer of A rectangular box with no top has...

Question 11

Find and classify the critical points of the following function: f(x, y) = + 30x³ - 15y³. A) saddle points are ( , - ) and (- , ), maximum at (3, 3) and minimum at (-3, -3) B) saddle points are (0, 0), ( , - ) and (- , ), minimum at (3, 3) and maximum at (-3, -3) C) saddle points are (0, 0), ( , - ), (- , ), (3, 3), and (-3, -3) D) saddle points are (0, 0), ( , - ) and (- , ), maximum at (3, 3) and minimum at (-3, -3) E) saddle points are (0, 0), ( , - ) and (- , ), maximum at (3, -3) and minimum at (-3, 3)

Accepted Answer

The answer of Find and classify the critical points of...

Question 12

Find and classify all critical points of f(x,y,z) = x³ + xz² + 3x² + y² + 2z² - 9x - 2y -10. A) local minimum at (-2, 1, 3), (-2, 1, -3), (-3, 1, 0) and saddle point at (1, 1, 0) B) local minimum at (-2, 1, 3), (-2, 1, -3) and local maximum at (-3, 1, 0), (1, 1, 0) C) local minimum at (1, 1, 0 ) and saddle point at (-2, 1, 3), (-2, 1, -3), (-3, 1, 0) D) local maximum at (-2, 1, 3), (-2, 1, -3), (-3, 1, 0) and saddle point at (1, 1, 0) E) local minimum at (-2, 1, 3) , (-3, 1, 0), local maximum at (1, 1, 0), and saddle point at (-2, 1, -3)

Accepted Answer

The answer of Find and classify all critical points of...

Question 13

If a function f(x,y) has a local or absolute extreme value at the point (x₀, y₀) in its domain, then (x₀, y₀) must be either a critical point of f, a singular point of f, or a boundary point of the domain of f.

Accepted Answer

For a function f(x, y) to have a local or absolute extreme value at a point, that point must either be a critical point (where the gradient of f is zero or undefined), a singular point (where the function or its derivatives are not defined), or a boundary point of the domain of f (where the function can't be further analyzed in all directions due to constraints of the domain).

Question 14

Suppose the critical points of f(x,y) = 2x³ - 15x² + 24x - y² - 6y + 1 occur at the points(1, -3) and (4, -3).Then the function f has: A) saddle points at (1, -3) and (4, -3) B) a local minimum at (4, - 3) and a saddle point at (1, -3) C) a local maximum at (1, - 3) and a saddle point at (4, - 3) D) a local maximum at (4, -3) and a saddle point at (1, -3) E) a local maximum at (4, -3) and a local minimum at (1, -3)

Accepted Answer

The answer of Suppose the critical points of f(x,y) =...

Question 15

Suppose that a function f(x,y) has a critical point  (a, b) at an interior point in its domain and that f has continuous second order partials in a neighbourhood of (a, b).
If     , then f has no local extremum at (a, b).

Accepted Answer

The answer of Suppose that a function f(x,y) has a...

Question 16

A closed rectangular container of volume 96 cubic metres is to be made from three different materials.The top and the bottom of the container are to be made from a material that costs $4 per square metre, two parallel sides (say left and right) are to be made from a material that costs $3 per square metre, and the other two parallel sides (front and back) are to be made from a material that costs $1 per square metre.Let x and y be the dimensions of the base of the container and z be its height in metres.(i) Express the total cost of the container (in dollars) as a function of x and y.(ii) Find dimensions of the most economical container and how much it costs.

Accepted Answer

The answer of A closed rectangular container of volume 96...

Question 17

Find the absolute maximum and minimum values of f(x, y) = y² - y - x - 5 on the square0 $\le$x $\le$ 1, 0 $\le$y $\le$ 1. A) maximum 0, minimum - B) maximum - , minimum - C) maximum -5, minimum - D) maximum -5, minimum - E) maximum - , minimum -5

Accepted Answer

The answer of Find the absolute maximum and minimum values...

Question 18

Find the absolute maximum and minimum of f(x, y) = 4x² + 2xy - 3y² on the unit square0 $\le$ x $\le$ 1, 0 $\le$y$\le$ 1. A) maximum , minimum -3 B) maximum 4, minimum -3 C) maximum 4, minimum 0 D) maximum , minimum -4 E) maximum , minimum 0

Accepted Answer

The answer of Find the absolute maximum and minimum of...

Question 19

Find the absolute maximum and minimum values of f(x, y) = 4(x - x²) sin($\pi$y) on the rectangle 0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 2 and the points where they are assumed. A) maximum 1 at , minimum 0 at B) maximum 2 at , minimum -2 at C) maximum 1 at , minimum -1 at D) maximum 1 at , minimum 0 at (0, 0), (1, 0), (0, 2), and (1, 2) E) maximum 1 at , minimum 0 at

Accepted Answer

The answer of Find the absolute maximum and minimum values...

Question 20

Find the absolute maximum and minimum values of the linear function f(x, y) = -2x + y - 10 on the polygon 0 $\le$ x $\le$2, 0 $\le$ y $\le$ 2, y - x $\le$ 1.
A) maximum -9, minimum -14
B) maximum -9, minimum -12
C) maximum -8, minimum -12
D) maximum -8, minimum -15
E) maximum -9, minimum -10

Accepted Answer

The absolute maximum and minimum values of a linear function on a polygon can be found by evaluating the function at the vertices of the polygon. The vertices of the given polygon are (0,0), (0,1), (1,2), and (2,1). Evaluating f(x, y) at these points gives: f(0,0) = -10, f(0,1) = -9, f(1,2) = -10, and f(2,1) = -14. Therefore, the maximum value is -9 and the minimum value is -14.

Quiz 14: Applications of Partial Derivatives

Find the critical points of f(x, y) = 3x² - 2xy + 2y² - 10x + 1.

Find the critical points of f(x, y) = ln (x² + y² + 4x - 4y + 8) .

Find all critical points of f(x) = 2x³y -4x³ + 6y³ -18y + 19.

Find and classify the critical points of the function f(x, y) = x sin y.

Find and classify the critical points of the function f(x, y) = 2y³ - 3x² - 3xy + 9x.

Find and classify all critical points for the function f(x, y) = .

Find and classify all critical points for the function f(x, y) = x³ - 12xy² + y³ + 45y.

Find the maximum and minimum values of f(x, y) = (x - y) .

Find and classify all critical points for the function f(x, y) = 2y³ + 3y² - 12y -x² + 2x.

A rectangular box with no top has given surface area S. Find the length, width, and height of the box if the volume is as large as possible.

Find and classify the critical points of the following function: f(x, y) = + 30x³ - 15y³.

Find and classify all critical points of f(x,y,z) = x³ + xz² + 3x² + y² + 2z² - 9x - 2y -10.

If a function f(x,y) has a local or absolute extreme value at the point (x₀, y₀) in its domain, then (x₀, y₀) must be either a critical point of f, a singular point of f, or a boundary point of the domain of f.

Suppose the critical points of f(x,y) = 2x³ - 15x² + 24x - y² - 6y + 1 occur at the points(1, -3) and (4, -3) .Then the function f has:

Suppose that a function f(x,y) has a critical point (a, b) at an interior point in its domain and that f has continuous second order partials in a neighbourhood of (a, b). If , then f has no local extremum at (a, b).

Find the absolute maximum and minimum values of f(x, y) = y² - y - x - 5 on the square0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 1.

Find the absolute maximum and minimum of f(x, y) = 4x² + 2xy - 3y² on the unit square0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 1.

Find the absolute maximum and minimum values of f(x, y) = 4(x - x²) sin( $\pi$ y) on the rectangle 0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 2 and the points where they are assumed.

Find the absolute maximum and minimum values of the linear function f(x, y) = -2x + y - 10 on the polygon 0 $\le$ x $\le$ 2, 0 $\le$ y $\le$ 2, y - x $\le$ 1.

Quiz 14: Applications of Partial Derivatives

Find the critical points of f(x, y) = 3x2 - 2xy + 2y2 - 10x + 1.

Find the critical points of f(x, y) = ln (x2 + y2 + 4x - 4y + 8) .

Find all critical points of f(x) = 2x3y -4x3 + 6y3 -18y + 19.

Find and classify the critical points of the function f(x, y) = x sin y.

Find and classify the critical points of the function f(x, y) = 2y3 - 3x2 - 3xy + 9x.

Find and classify all critical points for the function f(x, y) = .

Find and classify all critical points for the function f(x, y) = x3 - 12xy2 + y3 + 45y.

Find the maximum and minimum values of f(x, y) = (x - y) .

Find and classify all critical points for the function f(x, y) = 2y3 + 3y2 - 12y -x2 + 2x.

A rectangular box with no top has given surface area S. Find the length, width, and height of the box if the volume is as large as possible.

Find and classify the critical points of the following function: f(x, y) = + 30x3 - 15y3.

Find and classify all critical points of f(x,y,z) = x3 + xz2 + 3x2 + y2 + 2z2 - 9x - 2y -10.

If a function f(x,y) has a local or absolute extreme value at the point (x0, y0) in its domain, then (x0, y0) must be either a critical point of f, a singular point of f, or a boundary point of the domain of f.

Suppose the critical points of f(x,y) = 2x3 - 15x2 + 24x - y2 - 6y + 1 occur at the points(1, -3) and (4, -3) .Then the function f has:

Suppose that a function f(x,y) has a critical point (a, b) at an interior point in its domain and that f has continuous second order partials in a neighbourhood of (a, b). If , then f has no local extremum at (a, b).

Find the absolute maximum and minimum values of f(x, y) = y2 - y - x - 5 on the square0 ≤\le≤ x ≤\le≤ 1, 0 ≤\le≤ y ≤\le≤ 1.

Find the absolute maximum and minimum of f(x, y) = 4x2 + 2xy - 3y2 on the unit square0 ≤\le≤ x ≤\le≤ 1, 0 ≤\le≤ y ≤\le≤ 1.

Find the absolute maximum and minimum values of f(x, y) = 4(x - x2) sin( π\piπ y) on the rectangle 0 ≤\le≤ x ≤\le≤ 1, 0 ≤\le≤ y ≤\le≤ 2 and the points where they are assumed.

Find the absolute maximum and minimum values of the linear function f(x, y) = -2x + y - 10 on the polygon 0 ≤\le≤ x ≤\le≤ 2, 0 ≤\le≤ y ≤\le≤ 2, y - x ≤\le≤ 1.

Find the critical points of f(x, y) = 3x² - 2xy + 2y² - 10x + 1.

Find the critical points of f(x, y) = ln (x² + y² + 4x - 4y + 8) .

Find all critical points of f(x) = 2x³y -4x³ + 6y³ -18y + 19.

Find and classify the critical points of the function f(x, y) = 2y³ - 3x² - 3xy + 9x.

Find and classify all critical points for the function f(x, y) = x³ - 12xy² + y³ + 45y.

Find and classify all critical points for the function f(x, y) = 2y³ + 3y² - 12y -x² + 2x.

Find and classify the critical points of the following function: f(x, y) = + 30x³ - 15y³.

Find and classify all critical points of f(x,y,z) = x³ + xz² + 3x² + y² + 2z² - 9x - 2y -10.

If a function f(x,y) has a local or absolute extreme value at the point (x₀, y₀) in its domain, then (x₀, y₀) must be either a critical point of f, a singular point of f, or a boundary point of the domain of f.

Suppose the critical points of f(x,y) = 2x³ - 15x² + 24x - y² - 6y + 1 occur at the points(1, -3) and (4, -3) .Then the function f has:

Find the absolute maximum and minimum values of f(x, y) = y² - y - x - 5 on the square0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 1.

Find the absolute maximum and minimum of f(x, y) = 4x² + 2xy - 3y² on the unit square0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 1.

Find the absolute maximum and minimum values of f(x, y) = 4(x - x²) sin( $\pi$ y) on the rectangle 0 $\le$ x $\le$ 1, 0 $\le$ y $\le$ 2 and the points where they are assumed.

Find the absolute maximum and minimum values of the linear function f(x, y) = -2x + y - 10 on the polygon 0 $\le$ x $\le$ 2, 0 $\le$ y $\le$ 2, y - x $\le$ 1.