Lagrange Multipliers

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Use the method of Lagrange multipliers to find the minimum value of \(f(x,y)={x}^{2}+4{y}^{2}-2x+8y\) subject to the constraint \(x+2y=7.\)

   Ҷавоби ҷавобро нишон диҳед

   Let’s follow the problem-solving strategy:

   1. The optimization function is \(f(x,y)={x}^{2}+4{y}^{2}-2x+8y.\) To determine the constraint function, we must first subtract \(7\) from both sides of the constraint. This gives \(x+2y-7=0.\) The constraint function is equal to the left-hand side, so \(g(x,y)=x+2y-7.\) The problem asks us to solve for the minimum value of \(f,\) subject to the constraint (see the following graph).
      
   2. We then must calculate the gradients of both f and g:
      \[\begin{array}{l}∇f(x,y)=(2x-2)i+(8y+8)j \\ ∇g(x,y)=i+2j.\end{array}\]
      The equation \(∇f({x}_{0},{y}_{0})=\lambda ∇g({x}_{0},{y}_{0})\) becomes
      \[(2{x}_{0}-2)i+(8{y}_{0}+8)j=\lambda (i+2j),\]
      which can be rewritten as
      \[(2{x}_{0}-2)i+(8{y}_{0}+8)j=\lambda i+2\lambda j.\]
      Next, we set the coefficients of \(i\ \text{and}\ j\) equal to each other:
      \[\begin{array}{l}2{x}_{0}-2=\lambda \\ 8{y}_{0}+8=2\lambda .\end{array}\]
      The equation \(g({x}_{0},{y}_{0})=0\) becomes \({x}_{0}+2{y}_{0}-7=0.\) Therefore, the system of equations that needs to be solved is
      \[\begin{array}{lll}2{x}_{0}-2 & = & \lambda \\ 8{y}_{0}+8 & = & 2\lambda \\ {x}_{0}+2{y}_{0}-7 & = & 0.\end{array}\]
   3. This is a linear system of three equations in three variables. We start by solving the second equation for \(\lambda\) and substituting it into the first equation. This gives \(\lambda =4{y}_{0}+4,\) so substituting this into the first equation gives
      \[2{x}_{0}-2=4{y}_{0}+4.\]
      Solving this equation for \({x}_{0}\) gives \({x}_{0}=2{y}_{0}+3.\) We then substitute this into the third equation: \[\begin{array}{lll}(2{y}_{0}+3)+2{y}_{0}-7 & = & 0 \\ 4{y}_{0}-4 & = & 0 \\ {y}_{0} & = & 1.\end{array}\]
      Since \({x}_{0}=2{y}_{0}+3,\) this gives \({x}_{0}=5.\)
   4. Next, we substitute \((5,1)\) into \(f(x,y)={x}^{2}+4{y}^{2}-2x+8y,\) gives \(f(5,1)={5}^{2}+4{(1)}^{2}-2(5)+8(1)=27.\) To ensure this corresponds to a minimum value on the constraint function, let’s try some other values, such as the intercepts of \(g(x,y)=0,\) Which are \((7,0)\) and \((0,3.5).\) We get \(f(7,0)=35\) and \(f(0,3.5)=77,\) so it appears \(f\) has a minimum at \((5,1).\)

2. Use the method of Lagrange multipliers to find the maximum value of \(f(x,y)=9{x}^{2}+36xy-4{y}^{2}-18x-8y\) subject to the constraint \(3x+4y=32.\)

   Ҷавоби ҷавобро нишон диҳед

   \(f\) has a maximum value of \(976\) at the point \((8,2).\)

3. The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number \(x\) of golf balls sold per month (measured in thousands), and the number of hours per month of advertising y, according to the function

   \[z=f(x,y)=48x+96y-{x}^{2}-2xy-9{y}^{2},\]

   where \(z\) is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by \(20x+4y=216.\) Find the values of \(x\) and \(y\) that maximize profit, and find the maximum profit.

   Ҷавоби ҷавобро нишон диҳед

   Again, we follow the problem-solving strategy:

   1. The optimization function is \(f(x,y)=48x+96y-{x}^{2}-2xy-9{y}^{2}.\) To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by \(4,\) which gives \(5x+y-54=0.\) The constraint function is equal to the left-hand side, so \(g(x,y)=5x+y-54.\) The problem asks us to solve for the maximum value of \(f,\) subject to this constraint.
   2. So, we calculate the gradients of both \(f\ \text{and}\) \(g\text{:}\)
      \[\begin{array}{l}∇f(x,y)=(48-2x-2y)i+(96-2x-18y)j \\ ∇g(x,y)=5i+j.\end{array}\]
      The equation \(∇f({x}_{0},{y}_{0})=\lambda ∇g({x}_{0},{y}_{0})\) becomes
      \[(48-2{x}_{0}-2{y}_{0})i+(96-2{x}_{0}-18{y}_{0})j=\lambda (5i+j),\]
      which can be rewritten as
      \[(48-2{x}_{0}-2{y}_{0})i+(96-2{x}_{0}-18{y}_{0})j=\lambda 5i+\lambda j.\]
      We then set the coefficients of \(i\ \text{and}\ j\) equal to each other:
      \[\begin{array}{lll}48-2{x}_{0}-2{y}_{0} & = & 5\lambda \\ 96-2{x}_{0}-18{y}_{0} & = & \lambda .\end{array}\]
      The equation \(g({x}_{0},{y}_{0})=0\) becomes \(5{x}_{0}+{y}_{0}-54=0.\) Therefore, the system of equations that needs to be solved is
      \[\begin{array}{lll}48-2{x}_{0}-2{y}_{0} & = & 5\lambda \\ 96-2{x}_{0}-18{y}_{0} & = & \lambda \\ 5{x}_{0}+{y}_{0}-54 & = & 0.\end{array}\]
   3. We use the left-hand side of the second equation to replace \(\lambda\) in the first equation:
      \[\begin{array}{lll}48-2{x}_{0}-2{y}_{0} & = & 5(96-2{x}_{0}-18{y}_{0}) \\ 48-2{x}_{0}-2{y}_{0} & = & 480-10{x}_{0}-90{y}_{0} \\ 8{x}_{0} & = & 432-88{y}_{0} \\ {x}_{0} & = & 54-11{y}_{0}.\end{array}\]
      Then we substitute this into the third equation:
      \[\begin{array}{lll}5(54-11{y}_{0})+{y}_{0}-54 & = & 0 \\ 270-55{y}_{0}+{y}_{0} & = & 0 \\ 216-54{y}_{0} & = & 0 \\ {y}_{0} & = & 4.\end{array}\]
      Since \({x}_{0}=54-11{y}_{0},\) this gives \({x}_{0}=10.\)
   4. We then substitute \((10,4)\) into \(f(x,y)=48x+96y-{x}^{2}-2xy-9{y}^{2},\) which gives
      \[\begin{array}{ll}f(10,4) & =48(10)+96(4)-{(10)}^{2}-2(10)(4)-9{(4)}^{2} \\ & =480+384-100-80-144=540.\end{array}\]
      Therefore the maximum profit that can be attained, subject to budgetary constraints, is \(\text{\$}540,000\) with a production level of \(10,000\) golf balls and \(4\) hours of advertising bought per month. Let’s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are \((10.8,0)\) and \((0,54)\) Let’s evaluate \(f\) at both of these points:
      \[\begin{array}{lll}f(10.8,0) & = & 48(10.8)+96(0)-{10.8}^{2}-2(10.8)(0)-9({0}^{2})=401.76 \\ f(0,54) & = & 48(0)+96(54)-{0}^{2}-2(0)(54)-9({54}^{2})=-21,060.\end{array}\]
      The second value represents a loss, since no golf balls are produced. Neither of these values exceed \(540,\) so it seems that our extremum is a maximum value of \(f.\)

4. A company has determined that its production level is given by the Cobb-Douglas function \(f(x,y)=2.5{x}^{0.45}{y}^{0.55}\) where x represents the total number of labor hours in \(1\) year and y represents the total capital input for the company. Suppose \(1\) unit of labor costs \(\text{\$}40\) and \(1\) unit of capital costs \(\text{\$}50.\) Use the method of Lagrange multipliers to find the maximum value of \(f(x,y)=2.5{x}^{0.45}{y}^{0.55}\) subject to a budgetary constraint of \(\text{\$}500,000\) per year.

   Ҷавоби ҷавобро нишон диҳед

   A maximum production level of \(13890\) occurs with \(5625\) labor hours and \(\text{\$}5500\) of total capital input.

5. Find the minimum of the function \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}\) subject to the constraint \(x+y+z=1.\)

   Ҷавоби ҷавобро нишон диҳед

   1. The optimization function is \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}.\) To determine the constraint function, we subtract \(1\) from each side of the constraint: \(x+y+z-1=0\) which gives the constraint function as \(g(x,y,z)=x+y+z-1.\)
   2. Next, we calculate \(∇f(x,y,z)\) and \(∇g(x,y,z)\text{:}\)
      \[\begin{array}{l}∇f(x,y,z)=〈2x,2y,2z〉 \\ ∇g(x,y,z)=〈1,1,1〉\text{.}\end{array}\]
      This leads to the equations
      \[\begin{array}{lll}〈2{x}_{0},2{y}_{0},2{z}_{0}〉 & = & \lambda 〈1,1,1〉 \\ {x}_{0}+{y}_{0}+{z}_{0}-1 & = & 0\end{array}\]
      which can be rewritten in the following form:
      \[\begin{array}{lll}2{x}_{0} & = & \lambda \\ 2{y}_{0} & = & \lambda \\ 2{z}_{0} & = & \lambda \\ {x}_{0}+{y}_{0}+{z}_{0}-1 & = & 0.\end{array}\]
   3. Since each of the first three equations has \(\lambda\) on the right-hand side, we know that \(2{x}_{0}=2{y}_{0}=2{z}_{0}\) and all three variables are equal to each other. Substituting \({y}_{0}={x}_{0}\) and \({z}_{0}={x}_{0}\) into the last equation yields \(3{x}_{0}-1=0,\) so \({x}_{0}=\frac{1}{3}\) and \({y}_{0}=\frac{1}{3}\) and \({z}_{0}=\frac{1}{3}\) which corresponds to a critical point on the constraint curve.
   4. Then, we evaluate f at the point \((\frac{1}{3},\frac{1}{3},\frac{1}{3})\text{:}\)
      \[f(\frac{1}{3},\frac{1}{3},\frac{1}{3})={(\frac{1}{3})}^{2}+{(\frac{1}{3})}^{2}+{(\frac{1}{3})}^{2}=\frac{3}{9}=\frac{1}{3}.\]
      Therefore, an extremum of the function is \(\frac{1}{3}.\) To verify it is a minimum, choose other points that satisfy the constraint and calculate \(f\) at that point. For example,
      \[\begin{array}{lll}f(1,0,0) & = & {1}^{2}+{0}^{2}+{0}^{2}=1 \\ f(0,-2,3) & = & {0}^{2}+{(-2)}^{2}+{3}^{2}=13.\end{array}\]
      Both of these values are greater than \(\frac{1}{3},\) leading us to believe the extremum is a minimum.

6. Use the method of Lagrange multipliers to find the minimum value of the function

   \[f(x,y,z)=x+y+z\]

   subject to the constraint \({x}^{2}+{y}^{2}+{z}^{2}=1.\)

   Ҷавоби ҷавобро нишон диҳед

   \(\begin{array}{lll}f(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}) & = & \frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}+\frac{\sqrt{3}}{3}=\sqrt{3} \\ f(-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}) & = & -\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{3}=\text{-}\sqrt{3}.\end{array}\)

7. Find the local extreme values of the function

   \[f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}\]

   subject to the constraints \({z}^{2}={x}^{2}+{y}^{2}\) and \(x+y-z+1=0.\)

   Ҷавоби ҷавобро нишон диҳед

   Let’s follow the problem-solving strategy:

   1. The optimization function is \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}.\) To determine the constraint functions, we first subtract \({z}^{2}\) from both sides of the first constraint, which gives \({x}^{2}+{y}^{2}-{z}^{2}=0,\) so \(g(x,y,z)={x}^{2}+{y}^{2}-{z}^{2}.\) The second constraint function is \(h(x,y,z)=x+y-z+1.\)
   2. We then calculate the gradients of \(f,g,\ \text{and}\ h\text{:}\)
      \[\begin{array}{l}∇f(x,y,z)=2xi+2yj+2zk \\ ∇g(x,y,z)=2xi+2yj-2zk \\ ∇h(x,y,z)=i+j-k.\end{array}\]
      The equation \(∇f({x}_{0},{y}_{0},{z}_{0})={\lambda }_{1}∇g({x}_{0},{y}_{0},{z}_{0})+{\lambda }_{2}∇h({x}_{0},{y}_{0},{z}_{0})\) becomes
      \[2{x}_{0}i+2{y}_{0}j+2{z}_{0}k={\lambda }_{1}(2{x}_{0}i+2{y}_{0}j-2{z}_{0}k)+{\lambda }_{2}(i+j-k),\]
      which can be rewritten as
      \[2{x}_{0}i+2{y}_{0}j+2{z}_{0}k=(2{\lambda }_{1}{x}_{0}+{\lambda }_{2})i+(2{\lambda }_{1}{y}_{0}+{\lambda }_{2})j-(2{\lambda }_{1}{z}_{0}+{\lambda }_{2})k.\]
      Next, we set the coefficients of \(i\ \text{,}\ j\text{, and}\ k\) equal to each other:
      \[\begin{array}{l}2{x}_{0}=2{\lambda }_{1}{x}_{0}+{\lambda }_{2} \\ 2{y}_{0}=2{\lambda }_{1}{y}_{0}+{\lambda }_{2} \\ 2{z}_{0}=-2{\lambda }_{1}{z}_{0}-{\lambda }_{2}.\end{array}\]
      The two equations that arise from the constraints are \({z}_{0}{}^{2}={x}_{0}{}^{2}+{y}_{0}{}^{2}\) and \({x}_{0}+{y}_{0}-{z}_{0}+1=0.\) Combining these equations with the previous three equations gives
      \[\begin{array}{lll}2{x}_{0} & = & 2{\lambda }_{1}{x}_{0}+{\lambda }_{2} \\ 2{y}_{0} & = & 2{\lambda }_{1}{y}_{0}+{\lambda }_{2} \\ 2{z}_{0} & = & -2{\lambda }_{1}{z}_{0}-{\lambda }_{2} \\ {z}_{0}{}^{2} & = & {x}_{0}{}^{2}+{y}_{0}{}^{2} \\ {x}_{0}+{y}_{0}-{z}_{0}+1 & = & 0.\end{array}\]
   3. The first three equations contain the variable \({\lambda }_{2}.\) Solving the third equation for \({\lambda }_{2}\) and replacing into the first and second equations reduces the number of equations to four:
      \[\begin{array}{lll}2{x}_{0} & = & 2{\lambda }_{1}{x}_{0}-2{\lambda }_{1}{z}_{0}-2{z}_{0} \\ 2{y}_{0} & = & 2{\lambda }_{1}{y}_{0}-2{\lambda }_{1}{z}_{0}-2{z}_{0} \\ {z}_{0}{}^{2} & = & {x}_{0}{}^{2}+{y}_{0}{}^{2} \\ {x}_{0}+{y}_{0}-{z}_{0}+1 & = & 0.\end{array}\]
      Next, we solve the first and second equation for \({\lambda }_{1}.\) The first equation gives \({\lambda }_{1}=\frac{{x}_{0}+{z}_{0}}{{x}_{0}-{z}_{0}},\) the second equation gives \({\lambda }_{1}=\frac{{y}_{0}+{z}_{0}}{{y}_{0}-{z}_{0}}.\) We set the right-hand side of each equation equal to each other and cross-multiply:
      \[\begin{array}{lll}\frac{{x}_{0}+{z}_{0}}{{x}_{0}-{z}_{0}} & = & \frac{{y}_{0}+{z}_{0}}{{y}_{0}-{z}_{0}} \\ ({x}_{0}+{z}_{0})({y}_{0}-{z}_{0}) & = & ({x}_{0}-{z}_{0})({y}_{0}+{z}_{0}) \\ {x}_{0}{y}_{0}-{x}_{0}{z}_{0}+{y}_{0}{z}_{0}-{z}_{0}{}^{2} & = & {x}_{0}{y}_{0}+{x}_{0}{z}_{0}-{y}_{0}{z}_{0}-{z}_{0}2 \\ 2{y}_{0}{z}_{0}-2{x}_{0}{z}_{0} & = & 0 \\ 2{z}_{0}({y}_{0}-{x}_{0}) & = & 0.\end{array}.\]
      Therefore, either \({z}_{0}=0\) or \({y}_{0}={x}_{0}.\) If \({z}_{0}=0,\) then the first constraint becomes \(0={x}_{0}{}^{2}+{y}_{0}{}^{2}.\) The only real solution to this equation is \({x}_{0}=0\) and \({y}_{0}=0,\) which gives the ordered triple \((0,0,0).\) This point does not satisfy the second constraint, so it is not a solution.
      Next, we consider \({y}_{0}={x}_{0},\) which reduces the number of equations to three:
      \[\begin{array}{lll}{y}_{0} & = & {x}_{0} \\ {z}_{0}{}^{2} & = & {x}_{0}{}^{2}+{y}_{0}{}^{2} \\ {x}_{0}+{y}_{0}-{z}_{0}+1 & = & 0.\end{array}\]
      We substitute the first equation into the second and third equations:
      \[\begin{array}{lll}{z}_{0}{}^{2} & = & {x}_{0}{}^{2}+{x}_{0}{}^{2} \\ {x}_{0}+{x}_{0}-{z}_{0}+1 & = & 0.\end{array}\]
      Then, we solve the second equation for \({z}_{0},\) which gives \({z}_{0}=2{x}_{0}+1.\) We then substitute this into the first equation,
      \[\begin{array}{lll}{z}_{0}{}^{2} & = & 2{x}_{0}{}^{2} \\ {(2{x}_{0}+1)}^{2} & = & 2{x}_{0}{}^{2} \\ 4{x}_{0}{}^{2}+4{x}_{0}+1 & = & 2{x}_{0}{}^{2} \\ 2{x}_{0}{}^{2}+4{x}_{0}+1 & = & 0,\end{array}\]
      and use the quadratic formula to solve for \({x}_{0}\text{:}\)
      \[{x}_{0}=\frac{-4\pm \sqrt{{4}^{2}-4(2)(1)}}{2(2)}=\frac{-4\pm \sqrt{8}}{4}=\frac{-4\pm 2\sqrt{2}}{4}=-1\pm \frac{\sqrt{2}}{2}.\]
      Recall \({y}_{0}={x}_{0},\) so this solves for \({y}_{0}\) as well. Then, \({z}_{0}=2{x}_{0}+1,\) so
      \[{z}_{0}=2{x}_{0}+1=2(-1\pm \frac{\sqrt{2}}{2})+1=-2+1\pm \sqrt{2}=-1\pm \sqrt{2}.\]
      Therefore, there are two ordered triplet solutions:
      \[(-1+\frac{\sqrt{2}}{2},-1+\frac{\sqrt{2}}{2},-1+\sqrt{2})\ \text{and}\ (-1-\frac{\sqrt{2}}{2},-1-\frac{\sqrt{2}}{2},-1-\sqrt{2}).\]
   4. We substitute \((-1+\frac{\sqrt{2}}{2},-1+\frac{\sqrt{2}}{2},-1+\sqrt{2})\) into \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},\) which gives
      \[\begin{array}{ll}f(-1+\frac{\sqrt{2}}{2},-1+\frac{\sqrt{2}}{2},-1+\sqrt{2}) & ={(-1+\frac{\sqrt{2}}{2})}^{2}+{(-1+\frac{\sqrt{2}}{2})}^{2}+{(-1+\sqrt{2})}^{2} \\ & =(1-\sqrt{2}+\frac{1}{2})+(1-\sqrt{2}+\frac{1}{2})+(1-2\sqrt{2}+2) \\ & =6-4\sqrt{2}.\end{array}\]
      Then, we substitute \((-1-\frac{\sqrt{2}}{2},-1-\frac{\sqrt{2}}{2},-1-\sqrt{2})\) into \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},\) which gives
      \[\begin{array}{ll}f(-1-\frac{\sqrt{2}}{2},-1-\frac{\sqrt{2}}{2},-1-\sqrt{2}) & ={(-1-\frac{\sqrt{2}}{2})}^{2}+{(-1-\frac{\sqrt{2}}{2})}^{2}+{(-1-\sqrt{2})}^{2} \\ & =(1+\sqrt{2}+\frac{1}{2})+(1+\sqrt{2}+\frac{1}{2})+(1+2\sqrt{2}+2) \\ & =6+4\sqrt{2}.\end{array}\]
      \(6+4\sqrt{2}\) and \(6-4\sqrt{2}\) are the local extreme values of \(f(x,y,z),\) subject to the given constraints.

8. Use the method of Lagrange multipliers to find the minimum value of the function

   \[f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}\]

   subject to the constraints \(2x+y+2z=9\) and \(5x+5y+7z=29.\)

   Ҷавоби ҷавобро нишон диҳед

   \(f(2,1,2)=9\) is a minimum.

9. \(f(x,y)={x}^{2}y;{x}^{2}+2{y}^{2}=6\)

10. \(f(x,y,z)=xyz,{x}^{2}+2{y}^{2}+3{z}^{2}=6\)

    Ҷавоби ҷавобро нишон диҳед

    maximum: \(2\frac{\sqrt{3}}{3},\) minimum: \(\frac{-2\sqrt{3}}{3}\)

11. \(f(x,y)=xy;4{x}^{2}+8{y}^{2}=16\)

12. \(f(x,y)=4{x}^{3}+{y}^{2};2{x}^{2}+{y}^{2}=1\)

    Ҷавоби ҷавобро нишон диҳед

    maximum: \((\frac{\sqrt{2}}{2},0,\sqrt{2}),\) minimum: \((\frac{\text{-}\sqrt{2}}{2},0,\text{-}\sqrt{2})\)

13. \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},{x}^{4}+{y}^{4}+{z}^{4}=1\)

14. \(f(x,y,z)=yz+xy,xy=1,{y}^{2}+{z}^{2}=1\)

    Ҷавоби ҷавобро нишон диҳед

    maximum: \(\frac{3}{2},\) minimum = \(\frac{1}{2}\)

15. \(f(x,y)={x}^{2}+{y}^{2},{(x-1)}^{2}+4{y}^{2}=4\)

16. \(f(x,y)=4xy,\frac{{x}^{2}}{9}+\frac{{y}^{2}}{16}=1\)

    Ҷавоби ҷавобро нишон диҳед

    maxima: \(f(\frac{3\sqrt{2}}{2},2\sqrt{2})=24,\) \(f(-\frac{3\sqrt{2}}{2},-2\sqrt{2})=24;\) minima: \(f(-\frac{3\sqrt{2}}{2},2\sqrt{2})=-24,\) \(f(\frac{3\sqrt{2}}{2},-2\sqrt{2})=-24\)

17. \(f(x,y,z)=x+y+z,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

18. \(f(x,y,z)=x+3y-z,{x}^{2}+{y}^{2}+{z}^{2}=4\)

    Ҷавоби ҷавобро нишон диҳед

    maximum: \(2\sqrt{11}\) at \(f(\frac{2}{\sqrt{11}},\frac{6}{\sqrt{11}},\frac{-2}{\sqrt{11}});\) minimum: \(-2\sqrt{11}\) at \(f(\frac{-2}{\sqrt{11}},\frac{-6}{\sqrt{11}},\frac{2}{\sqrt{11}})\)

19. \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},xyz=4\)

20. Minimize \(f(x,y)={x}^{2}+{y}^{2}\) on the hyperbola \(xy=1.\)

    Ҷавоби ҷавобро нишон диҳед

    \(2.0\)

21. Minimize \(f(x,y)=xy\) on the ellipse \({b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}.\)

22. Maximize \(f(x,y,z)=2x+3y+5z\) on the sphere \({x}^{2}+{y}^{2}+{z}^{2}=19.\)

    Ҷавоби ҷавобро нишон диҳед

    \(19\sqrt{2}\)

23. Maximize \(\begin{array}{l}f(x,y)={x}^{2}-{y}^{2};x>0,y>0; \\ g(x,y)=y-{x}^{2}=0\end{array}\)

24. The curve \({x}^{3}-{y}^{3}=1\) is asymptotic to the line \(y=x.\) Find the point(s) on the curve \({x}^{3}-{y}^{3}=1\) farthest from the line \(y=x.\)

    Ҷавоби ҷавобро нишон диҳед

    \((\frac{1}{\sqrt[3]{2}},\frac{-1}{\sqrt[3]{2}})\)

25. Maximize \(U(x,y)=8{x}^{4\text{/}5}{y}^{1\text{/}5};4x+2y=12\)

26. Minimize \(f(x,y)={x}^{2}+{y}^{2},x+2y-5=0.\)

    Ҷавоби ҷавобро нишон диҳед

    \(f(1,2)=5\)

27. Maximize \(f(x,y)=\sqrt{6-{x}^{2}-{y}^{2}},x+y-2=0.\)

28. Minimize \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2},x+y+z=1.\)

    Ҷавоби ҷавобро нишон диҳед

    \(f(\frac{1}{3},\frac{1}{3},\frac{1}{3})=\frac{1}{3}\)

29. Minimize \(f(x,y)={x}^{2}-{y}^{2}\) subject to the constraint \(x-2y+6=0.\)

30. Minimize \(f(x,y,z)={x}^{2}+{y}^{2}+{z}^{2}\) when \(x+y+z=9\) and \(x+2y+3z=20.\)

    Ҷавоби ҷавобро нишон диҳед

    minimum: \(f(2,3,4)=29\)

31. A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is \(10\) in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

32. A rectangular box without a top (a topless box) is to be made from \(12\) ft² of cardboard. Find the maximum volume of such a box.

    Ҷавоби ҷавобро нишон диҳед

    The maximum volume is \(4\) ft³. The dimensions are \(1\ \times \ 2\ \times \ 2\) ft.

33. Find the minimum and maximum distances between the ellipse \({x}^{2}+xy+2{y}^{2}=1\) and the origin.

34. Find the point on the surface \({x}^{2}-2xy+{y}^{2}-x+y=0\) closest to the point \((1,2,-3).\)

    Ҷавоби ҷавобро нишон диҳед

    The point on the surface \({x}^{2}-2xy+{y}^{2}-x+y=0\) closest to the point \((1,2,-3)\) is \(\left(\frac{3}{2},\frac{3}{2},-3\right)\).

35. Show that, of all the triangles inscribed in a circle of radius \(R\) (see diagram), the equilateral triangle has the largest perimeter.

36. Find the minimum distance from point \((0,1)\) to the parabola \({x}^{2}=4y.\)

    Ҷавоби ҷавобро нишон диҳед

    \(1.0\)

37. Find the minimum distance from the parabola \(y={x}^{2}\) to point \((0,3).\)

38. Find the minimum distance from the plane \(x+y+z=1\) to point \((2,1,1).\)

    Ҷавоби ҷавобро нишон диҳед

    \(\sqrt{3}\)

39. A large container in the shape of a rectangular solid must have a volume of \(480\) m³. The bottom of the container costs $5/m² to construct whereas the top and sides cost $3/m² to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost.

40. Find the point on the line \(y=2x+3\) that is closest to point \((4,2).\)

    Ҷавоби ҷавобро нишон диҳед

    \((\frac{2}{5},\frac{19}{5})\)

Санҷиши худ

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

& Иловаи забон Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems