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

If \(\displaystyle f\) is a three-dimensional function, the Lagrangian function expands to three equations,

\(\displaystyle f_x=\lambda g_x\), \(\displaystyle f_y=\lambda g_y\) and \(\displaystyle f_z=\lambda g_z\).

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

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

\(\displaystyle f_x=yz\) , \(\displaystyle f_y=xz\), \(\displaystyle f_z=xy\), 

\(\displaystyle g_x=2y+2z\), \(\displaystyle g_y=2x+2z\), \(\displaystyle g_z=2x+2y\)

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

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

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

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

\(\displaystyle 2xy+2xz+2yz=125\)

We have four equations and four variables (\(\displaystyle x\),\(\displaystyle y\), \(\displaystyle z\) and \(\displaystyle \lambda\)), so we can solve the system of equations.

Multiplying the first equation by \(\displaystyle x\) and the second equation by \(\displaystyle y\) gives us

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

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

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

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

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

\(\displaystyle 2xy+2xz=2xy+2yz\)

\(\displaystyle 2xz=2yz\)

\(\displaystyle x=y\)

Multiplying the first equation by \(\displaystyle x\) and the second equation by \(\displaystyle z\) gives us

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

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

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

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

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

\(\displaystyle 2xy+2xz=2xz+2yz\)

\(\displaystyle 2xy=2yz\)

\(\displaystyle x=z\)

We now know \(\displaystyle x=y=z\). Substituting \(\displaystyle y=x\) and \(\displaystyle z=x\) into the constraint gives us

\(\displaystyle 2xy+2xz+2yz=125\)

\(\displaystyle 2x^2+2x^2+2x^2=125\)

\(\displaystyle 6x^2=125\)

\(\displaystyle x^2=20.833\Rightarrow x=4.56in\)

\(\displaystyle y=4.56in\)

\(\displaystyle z=4.56in\)

These dimensions maximize the volume of the box.

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

If \(\displaystyle f\) is a two-dimensional function, the Lagrangian function expands to two equations,

\(\displaystyle f_r=\lambda g_r\) and \(\displaystyle f_h=\lambda g_h\).

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

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

\(\displaystyle f_r=2\pi h+2\pi r\) , \(\displaystyle f_h=2\pi r\), \(\displaystyle g_r=2\pi rh\), \(\displaystyle g_h=\pi r^2\)

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

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

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

\(\displaystyle \pi r^2h=27\)

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

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

\(\displaystyle \frac{h+r}{ rh}=\lambda\)

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

\(\displaystyle \frac{2}{ r}= \lambda\)

Setting both expressions of lambda equal to each other gives us

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

\(\displaystyle r^2=rh\)

\(\displaystyle r=h\)

Substituting this expression into the constraint, we have

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

\(\displaystyle \pi r^3=27\Rightarrow r=2.05m\) 

\(\displaystyle h=2.05m\)

These dimensions minimize the surface area of the fish tank.

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

If \(\displaystyle f\) is a two-dimensional function, the Lagrangian function expands to two equations,

\(\displaystyle f_r=\lambda g_r\) and \(\displaystyle f_h=\lambda g_h\).

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

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

\(\displaystyle f_r=2\pi h+4\pi r\) , \(\displaystyle f_h=2\pi r\), \(\displaystyle g_r=2\pi rh\), \(\displaystyle g_h=\pi r^2\)

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

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

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

\(\displaystyle \pi r^2h=0.5\)

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

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

\(\displaystyle \frac{h+2r}{ rh}=\lambda\)

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

\(\displaystyle \frac{2}{ r}= \lambda\)

Setting both expressions of lambda equal to each other gives us

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

\(\displaystyle 2r^2=rh\)

\(\displaystyle 2r=h\)

Substituting this expression into the constraint, we have

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

\(\displaystyle 2\pi r^3=0.5\Rightarrow r=0.43dm\) 

\(\displaystyle h=2r=2*0.43=0.86dm\)

These dimensions minimize the surface area of the soda can.

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

If \(\displaystyle f\) is a two-dimensional function, the Lagrangian function expands to two equations,

\(\displaystyle f_x=\lambda g_x\) and \(\displaystyle f_y=\lambda g_y\).

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

The constraint is the area of the fence, or \(\displaystyle xy=72\).

\(\displaystyle f_x=2\) , \(\displaystyle f_y=1\), \(\displaystyle g_x=y\), \(\displaystyle g_y=x\)

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

\(\displaystyle 2=\lambda y\)

\(\displaystyle 1=\lambda x\)

\(\displaystyle xy=72\)

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

\(\displaystyle 2=\lambda y\Rightarrow \lambda=2/y\)

\(\displaystyle 1=\lambda x\Rightarrow \lambda=1/x\)

Setting the two expressions for \(\displaystyle \lambda\) equal to each other gives us

\(\displaystyle \frac{2}{y}=\frac{1}{x}\Rightarrow x=y/2\)

Substituting this expression into the constraint gives us

\(\displaystyle xy=72 \Rightarrow \frac{1}{2}y^2=72\)

\(\displaystyle y^2=144\)

\(\displaystyle y=12m\)

\(\displaystyle x=y/2=12/2=6m\)

These dimensions minimize the perimeter of the yard.

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

If \(\displaystyle f\) is a two-dimensional function, the Lagrangian function expands to two equations,

\(\displaystyle f_x=\lambda g_x\) and \(\displaystyle f_y=\lambda g_y\).

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

The constraint is the area of the box, or \(\displaystyle xy=16\).

\(\displaystyle f_x=2\) , \(\displaystyle f_y=2\), \(\displaystyle g_x=y\), \(\displaystyle g_y=x\)

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

\(\displaystyle 2=\lambda y\)

\(\displaystyle 2=\lambda x\)

\(\displaystyle xy=16\)

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

\(\displaystyle 2=\lambda y\Rightarrow \lambda=2/y\)

\(\displaystyle 2=\lambda x\Rightarrow \lambda=2/x\)

\(\displaystyle \frac{2}{y}=\frac{2}{x}\Rightarrow x=y\)

\(\displaystyle xy=16 \Rightarrow y^2=16\)

\(\displaystyle y=4m\)

\(\displaystyle x=4m\)

These dimensions minimize the perimeter of the sandbox.

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

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

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

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

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

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

Now lets apply this to our problem

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

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

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

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

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: 

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

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

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

x:

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

y:

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

z:

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

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: 

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

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

x:

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

y:

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

z:

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

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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

x:

\(\displaystyle \frac{\delta f}{\delta x}=6yz+2y+z=3\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=6xz+2x=60\)

z:

\(\displaystyle \frac{\delta f}{\delta z}=6xy+x=3\)

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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

x:

\(\displaystyle \frac{\delta f}{\delta x}=5yz=60\)

y:

\(\displaystyle \frac{\delta f}{\delta y}=5xz=40\)

z:

\(\displaystyle \frac{\delta f}{\delta z}=5xy=30\)