Answer:
4
Step-by-step explanation:
set
constrain:
Partial derivatives:
Lagrange multiplier:
![\left[\begin{array}{ccc}1\\1\end{array}\right]=a\left[\begin{array}{ccc}2x\\2y\end{array}\right]+b\left[\begin{array}{ccc}3x^2\\3y^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%5C%5C1%5Cend%7Barray%7D%5Cright%5D%3Da%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2x%5C%5C2y%5Cend%7Barray%7D%5Cright%5D%2Bb%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3x%5E2%5C%5C3y%5E2%5Cend%7Barray%7D%5Cright%5D)
4 equations:
By solving:
Second mathod:
Solve for x^2+y^2 = 7, x^3+y^3=10 first:
The maximum is 4
Answer:
A
Step-by-step explanation:
well some people like to go on a budget and others prefer not to because they rather spend their money money on wants rather than needs so I really hope this helps.
Answer:
y =14
11x = 11*3.5 =38.5
3x+2y = 11x = 38.5
Step-by-step explanation:
The perimeter of a triangle is the sum of all three sides
3x+2y + 11x+y = 91
Combine like terms
14x +3y = 91
The lines mean the sides are equal
3x+2y = 11x
Simplify
Subtract 3x from each side
3x-3x+2y = 11x-3x
2y = 8x
Divide by 2
2y/2 = 8x/2
y = 4x
Substitute 4x in the first equation every time you see y
14x +3y = 91
14x +3(4x) = 91
14x+12x=91
26x = 91
Divide by 26
26x/26 =91/26
x = 3.5
Now we can find y
y = 4x
y = 4(3.5)
y = 14
We know x and y we can find the length of each of the sides
y =14
11x = 11*3.5 =38.5
3x+2y = 11x = 38.5
Answer:
-94%, -0.925, -9/10, -8/9
Step-by-step explanation:
-94% = -0.94
-8/9 = -0.89
-9/10 = -0.90
Since the values are negative the order is opposite then if they were all positive.
Answer:
The solution to this question can be defined as follows:
Step-by-step explanation:
Please find the complete question in the attached file.
![A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D)
now for given values:
![\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%20-%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%3D0%20%5C%5C%5C%5C)
![\to (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} - \frac{1}{4}] =0 \\\\\to (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\](https://tex.z-dn.net/?f=%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B-%5Clambda%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%20-0%5D%20-%200%20-%20%5Cfrac%7B1%7D%7B4%7D%5B0-%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B%28%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Clambda%5E2%20%29%5D%20-%20%5Cfrac%7B1%7D%7B4%7D%5B%5Cfrac%7B%5Clambda%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7D%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B8%7D%5Clambda%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%5Clambda%5E2%20-%20%5Cfrac%7B%5Clambda%5E2%7D%7B2%7D%20-%20%5Clambda%5E3%20-%20%5Cfrac%7B%5Clambda%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%3D0%20%5C%5C%5C%5C%5Cto%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%28%5Clambda%20-%5Cfrac%7B1%7D%7B4%7D%29%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%3D0%5C%5C%5C%5C)
In point b:
Its
spectral radius is less than 1 hence matrix is convergent.
In point c:
![\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) = \left(\begin{array}{c}3&1&2\end{array}\right) , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\ \to x^{(k+1)} = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right] \\\\](https://tex.z-dn.net/?f=%5Cto%20c%5E%7B%28k%2B1%29%7D%20%3D%20A%20x%5E%7Bk%7D%2BC%20%5C%5C%5C%5C%5Cto%20x%280%29%20%3D%20%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D3%261%262%5Cend%7Barray%7D%5Cright%29%20%20%2C%20c%20%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%29%5C%5C%5C%5C%20%20%5Cto%20x%5E%7B%28k%2B1%29%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D%20x%5Ek%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%5C%5C)
after solving the value the answer is
:
![\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Clim_%7Bk%20%5Cto%20%5Cinfty%7D%20x%5Ek%3Do%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
