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Calculus 3 : Applications of Partial Derivatives

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Example Questions

Example Question #11 : Lagrange Multipliers

A box has a surface area of 125in^3 . What length, width and height maximize the volume of the box?

Possible Answers:

x=3.71in , y=5.93in , z=9.55in

x=4.56in , y=4.56in , z=4.56in

x=8.12in , y=8.12in , z=11.2in

Correct answer:

x=4.56in , y=4.56in , z=4.56in

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a three-dimensional function, the Lagrangian function expands to three equations,

$f_x=\lambda g_x$ , $f_y=\lambda g_y$ and $f_z=\lambda g_z$ .

In this problem, we are trying to maximize the volume of the box, so the equation being optimized is f=xyz .

The constraint is the surface area of the box, or 2xy+2xz+2yz=125 .

f_x=yz , f_y=xz , f_z=xy ,

g_x=2y+2z , g_y=2x+2z , g_z=2x+2y

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$yz=\lambda (2y+2z)$

$xz=\lambda (2x+2z)$

$xy=\lambda (2x+2y)$

2xy+2xz+2yz=125

We have four equations and four variables (,, and $\lambda$ ), so we can solve the system of equations.

Multiplying the first equation by and the second equation by gives us

$xyz=\lambda x(2y+2z)$

$xyz=\lambda y(2x+2z)$

The left side of both equations are the same, so we can set the right sides equal to each other

$\lambda x(2y+2z)=\lambda y(2x+2z)$

x(2y+2z)=y(2x+2z)

2xy+2xz=2xy+2yz

2xz=2yz

x=y

Multiplying the first equation by and the second equation by gives us

$xyz=\lambda x(2y+2z)$

$xyz=\lambda z(2x+2y)$

The left side of both equations are the same, so we can set the right sides equal to each other

$\lambda x(2y+2z)=\lambda z(2x+2y)$

x(2y+2z)=z(2x+2y)

2xy+2xz=2xz+2yz

2xy=2yz

x=z

We now know x=y=z . Substituting y=x and z=x into the constraint gives us

2xy+2xz+2yz=125

2x^2+2x^2+2x^2=125

6x^2=125

$x^2=20.833\Rightarrow x=4.56in$

y=4.56in

z=4.56in

These dimensions maximize the volume of the box.

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Example Question #11 : Lagrange Multipliers

A fish tank (right cylinder) with no top has a volume of 27m^3 . What height and radius will minimize the surface area of the fish tank?

Possible Answers:

r=12.4m , h=5.12m

r=2.05m , h=2.05m

r=4.36m , h=4.36m

r=9m , h=3m

Correct answer:

r=2.05m , h=2.05m

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_r=\lambda g_r$ and $f_h=\lambda g_h$ .

In this problem, we are trying to minimize the surface area of the fish tank with no top, so the equation being optimized is $f=2\pi rh+\pi r^2$ .

The constraint is the volume of the cylinder, or $\pi r^2h=27$ .

$f_r=2\pi h+2\pi r$ , $f_h=2\pi r$ , $g_r=2\pi rh$ , $g_h=\pi r^2$

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2\pi h+2\pi r=\lambda\left ( 2\pi rh\right )$

$2\pi r=\lambda \left (\pi r^2 \right )$

$\pi r^2h=27$

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2\pi h+2\pi r=\lambda\left ( 2\pi rh\right )\Rightarrow \frac{2\pi h+2\pi r}{ 2\pi rh}=\lambda$

$\frac{h+r}{ rh}=\lambda$

$2\pi r=\lambda \left (\pi r^2 \right )\Rightarrow \frac{2\pi r}{\pi r^2}= \lambda$

$\frac{2}{ r}= \lambda$

Setting both expressions of lambda equal to each other gives us

$\frac{h+r}{ rh}=\frac{2}{r}\Rightarrow r(h+r)=2rh\Rightarrow rh+r^2=2rh$

r^2=rh

r=h

Substituting this expression into the constraint, we have

$\pi r^2h=27\Rightarrow \pi r^2(r)=27$

$\pi r^3=27\Rightarrow r=2.05m$

h=2.05m

These dimensions minimize the surface area of the fish tank.

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Example Question #11 : Lagrange Multipliers

A soda can (a right cylinder) has a volume of 0.5dm^3 . What height and radius will minimize the surface area of the soda can?

Possible Answers:

r=0.27dm , h=0.49dm

r=0.39dm , h=0.56dm

r=0.57dm , h=0.94dm

r=0.43dm , h=0.86dm

Correct answer:

r=0.43dm , h=0.86dm

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_r=\lambda g_r$ and $f_h=\lambda g_h$ .

In this problem, we are trying to minimize the surface area of the soda can, so the equation being optimized is $f=2\pi rh+2\pi r^2$ .

The constraint is the volume of the cylinder, or $\pi r^2h=0.5$ .

$f_r=2\pi h+4\pi r$ , $f_h=2\pi r$ , $g_r=2\pi rh$ , $g_h=\pi r^2$

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2\pi h+4\pi r=\lambda\left ( 2\pi rh\right )$

$2\pi r=\lambda \left (\pi r^2 \right )$

$\pi r^2h=0.5$

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2\pi h+4\pi r=\lambda\left ( 2\pi rh\right )\Rightarrow \frac{2\pi h+4\pi r}{ 2\pi rh}=\lambda$

$\frac{h+2r}{ rh}=\lambda$

$2\pi r=\lambda \left (\pi r^2 \right )\Rightarrow \frac{2\pi r}{\pi r^2}= \lambda$

$\frac{2}{ r}= \lambda$

Setting both expressions of lambda equal to each other gives us

$\frac{h+2r}{ rh}=\frac{2}{r}\Rightarrow r(h+2r)=2rh\Rightarrow rh+2r^2=2rh$

2r^2=rh

2r=h

Substituting this expression into the constraint, we have

$\pi r^2h=0.5\Rightarrow \pi r^2(2r)=0.5$

$2\pi r^3=0.5\Rightarrow r=0.43dm$

h=2r=2*0.43=0.86dm

These dimensions minimize the surface area of the soda can.

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Example Question #14 : Lagrange Multipliers

What is the least amount of fence required to make a yard bordered on one side by a house? The area of the yard is 72ft^2 .

Possible Answers:

x=18m , y=4m

x=8m , y=9m

x=6m , y=12m

x=9m , y=8m

Correct answer:

x=6m , y=12m

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

In this problem, we are trying to minimize the perimeter of the yard, which is three sides, so the equation being optimized is f=2x+y .

The constraint is the area of the fence, or xy=72 .

f_x=2 , f_y=1 , g_x=y , g_y=x

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2=\lambda y$

$1=\lambda x$

xy=72

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2=\lambda y\Rightarrow \lambda=2/y$

$1=\lambda x\Rightarrow \lambda=1/x$

Setting the two expressions for $\lambda$ equal to each other gives us

$\frac{2}{y}=\frac{1}{x}\Rightarrow x=y/2$

Substituting this expression into the constraint gives us

$xy=72 \Rightarrow \frac{1}{2}y^2=72$

y^2=144

y=12m

x=y/2=12/2=6m

These dimensions minimize the perimeter of the yard.

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Example Question #12 : Lagrange Multipliers

What is the least amount of wood required to make a rectangular sandbox whose area is 16m^2 ?

Possible Answers:

x=8m , y=2m

x=4m , y=4m

x=16m , y=1m

x=2m , y=8m

Correct answer:

x=4m , y=4m

Explanation:

To optimize a function subject to the constraint g=k , we use the Lagrangian function, $\bigtriangledown f=\lambda \bigtriangledown g$ , where $\lambda$ is the Lagrangian multiplier.

If is a two-dimensional function, the Lagrangian function expands to two equations,

$f_x=\lambda g_x$ and $f_y=\lambda g_y$ .

In this problem, we are trying to minimize the perimeter of the sandbox, so the equation being optimized is f=2x+2y .

The constraint is the area of the box, or xy=16 .

f_x=2 , f_y=2 , g_x=y , g_y=x

Substituting these variables into the the Lagrangian function and the constraint equation gives us the following equations

$2=\lambda y$

$2=\lambda x$

xy=16

We have three equations and three variables (,, and $\lambda$ ), so we can solve the system of equations.

$2=\lambda y\Rightarrow \lambda=2/y$

$2=\lambda x\Rightarrow \lambda=2/x$

$\frac{2}{y}=\frac{2}{x}\Rightarrow x=y$

$xy=16 \Rightarrow y^2=16$

y=4m

x=4m

These dimensions minimize the perimeter of the sandbox.

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ at z=(0,5) .

Possible Answers:

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to , and , and plug in z=(1,5) .

$f(x,y)=\ln(4x^3+10y^2)$ , $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ , $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ , $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=xsin(y)cos(z) at the point $(\pi,\pi,\pi)$

Possible Answers:

$(0,\pi,0)$

$(\pi,-\pi,0)$

$(\pi,-\pi,-\pi)$

$(\pi,\pi,0)$

$(0,-\pi,0)$

Correct answer:

$(0,\pi,0)$

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

d[sin(u)]=cos(u)du

d[cos(u)]=-sin(u)du

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y,z)=xsin(y)cos(z) at the point $(\pi,\pi,\pi)$

$\frac{\delta f}{\delta x}=sin(y)cos(y)=0$

$\frac{\delta f}{\delta y}=xcos(y)cos(z)=\pi$

$\frac{\delta f}{\delta z}=-xsin(y)cos(y)=0$

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Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=xy^2sin(z) at the point $(3,2,2\pi)$

Possible Answers:

(4,4,4)

(4,12,12)

(0,0,0)

(4,12,0)

(0,0,12)

Correct answer:

(0,0,12)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y,z)=xy^2sin(z) at the point $(3,2,2\pi)$

$\frac{\delta f}{\delta x}=y^2sin(z)=0$

$\frac{\delta f}{\delta y}=2xysin(z)=0$

$\frac{\delta f}{\delta z}=xy^2cos(z)=12$

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Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=6xyz+2xy+xz at the point (3,0,3)

Possible Answers:

(3,60,21)

(18,0,18)

(21,60,21)

(3,60,3)

(18,6,3)

Correct answer:

(3,60,3)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y,z)=6xyz+2xy+xz at the point (3,0,3)

$\frac{\delta f}{\delta x}=6yz+2y+z=3$

$\frac{\delta f}{\delta y}=6xz+2x=60$

$\frac{\delta f}{\delta z}=6xy+x=3$

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Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

Find the slope of the function f(x,y,z)=5xyz at the point (2,3,4)

Possible Answers:

(30,40,60)

(40,30,60)

(40,60,30)

(60,30,40)

(60,40,30)

Correct answer:

(60,40,30)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point.

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Looking at f(x,y,z)=5xyz at the point (2,3,4)

$\frac{\delta f}{\delta x}=5yz=60$

$\frac{\delta f}{\delta y}=5xz=40$

$\frac{\delta f}{\delta z}=5xy=30$

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Calculus 3 : Applications of Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Lagrange Multipliers

Example Question #11 : Lagrange Multipliers

Example Question #11 : Lagrange Multipliers

Example Question #14 : Lagrange Multipliers

Example Question #12 : Lagrange Multipliers

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #1 : Gradient Vector, Tangent Planes, And Normal Lines

Example Question #2 : Gradient Vector, Tangent Planes, And Normal Lines

All Calculus 3 Resources

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