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11Alexandr11 [23.1K]
3 years ago
10

Find the greatest solution for x+y when x^2+y^2 = 7, x^3+y^3=10

Mathematics
1 answer:
damaskus [11]3 years ago
6 0

Answer:

4

Step-by-step explanation:

set

f(x,y)=x+y\\

constrain:

g(x,y)=x^2+y^2 = 7\\h(x,y)=x^3+y^3=10

Partial derivatives:

f_{x}=1\\f_{y} =1 \\g_{x}=2x \\g_{y}=2y\\h_{x}=3x^2 \\h_{y}=3y^2

Lagrange multiplier:

grad(f)=a*grad(g)+b*grad(h)\\

\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]

4 equations:

1=2ax+3bx^2\\1=2ay+3by^2\\x^2+y^2=7\\x^3+y^3=10

By solving:

a=4/9\\b=-2/27\\x+y=4

Second mathod:

Solve for x^2+y^2 = 7, x^3+y^3=10 first:

x=\frac{1}{2} -\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} +\frac{\sqrt{13}}{2} \\x=\frac{1}{2} +\frac{\sqrt{13}}{2} \ or \ y=\frac{1}{2} -\frac{\sqrt{13}}{2} \\x+y=-5\ or\ 1 \or\ 4

The maximum is 4

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