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denis-greek [22]
3 years ago
7

Find the minimum and maximum values of the function subject to the given constraint. (if an answer does not exist, enter dne.) f

(x, y) = x2y + x + y, xy = 5
Mathematics
1 answer:
nata0808 [166]3 years ago
8 0
Via Lagrange multipliers:

L(x,y,\lambd)=x^2y+x+y+\lambda(xy-5)
L_x=2xy+1+\lambda y=0
L_y=x^2+1+\lambda x=0
L_\lambda=xy-5=0

\underbrace{10}_{2xy}+1+\lambda y=0\implies \lambda=-\dfrac{11}y
xy=5\implies y=\dfrac5x\implies\lambda=-\dfrac{11}5x

\impliesx^2+1+\left(-\dfrac{11}5x\right)x=0\implies x^2=\dfrac56\implies x=\pm\sqrt{\dfrac56}
xy=5\implies y=\pm\sqrt{30}

At these points, we get local minima of f\left(\pm\sqrt{\dfrac56},\pm\sqrt{30}\right)=\pm2\sqrt{30}.

- - -

Another way to do this is to make f(x,y) a function independent of y, which is made possible by the constraint.

xy=5\implies y=\dfrac5x
\implies f(x,y)=f\left(x,\dfrac5x\right)=F(x)=6x+\dfrac5x
\implies F'(x)=6-\dfrac5{x^2}=0\implies x=\pm\sqrt{\dfrac56}

and so on.
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