(a) Find the maximum and minimum values of the function f(x, y, z) = 3x + 6y + 2z subject to the constraint 2x^2+4y^2+z^2=70. (b) Consider f (x, y) = xye^x-y (i) Find the rate of change of the function f(x,y) at the point P(5,5) in the direction of the vector u=(1,2). (ii) In which direction does the functionf(x, y) increase most rapidly at the point P(5,5)? (iii) What is the maximum rate of change off(x,y) at the point P(5,5)? (c ) Let f(x, y) = x^2 + 2y^2 - x^2y + 5. Find all critical points of f(x,y) and determine their nature (local max or local min or saddle point).

Question-Answer › Category: Mathematics › (a) Find the maximum and minimum values of the function f(x, y, z) = 3x + 6y + 2z subject to the constraint 2x^2+4y^2+z^2=70. (b) Consider f (x, y) = xye^x-y (i) Find the rate of change of the function f(x,y) at the point P(5,5) in the direction of the vector u=(1,2). (ii) In which direction does the functionf(x, y) increase most rapidly at the point P(5,5)? (iii) What is the maximum rate of change off(x,y) at the point P(5,5)? (c ) Let f(x, y) = x^2 + 2y^2 – x^2y + 5. Find all critical points of f(x,y) and determine their nature (local max or local min or saddle point).

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S gill asked 1 year ago

(a) Find the maximum and minimum values of the function f(x, y, z) = 3x + 6y + 2z subject to the constraint 2x^2+4y^2+z^2=70. (b) Consider f (x, y) = xye^x-y (i) Find the rate of change of the function f(x,y) at the point P(5,5) in the direction of the vector u=(1,2). (ii) In which direction does the functionf(x, y) increase most rapidly at the point P(5,5)? (iii) What is the maximum rate of change off(x,y) at the point P(5,5)? (c ) Let f(x, y) = x^2 + 2y^2 – x^2y + 5. Find all critical points of f(x,y) and determine their nature (local max or local min or saddle point).

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Richards answered 1 year ago

(a)
$f(x,y,z) = 3x+6y+2z \implies f_x = 3 \ ; \ f_y = 6 \ ; \ f_z = 2$
subject to the constraint:
$g(x,y,z) = 2x^2+4y^2+z^2-70=0 \implies g_x = 4x \ ; \ g_y = 8y \ ; \ g_z = 2z$
Using Lagrange Multiplier $\lambda$ , we get the following equations by Lagrange Multiplier Method:
$3 = 4\lambda x \ ; \ 6 = 8\lambda y \implies 3 = 4\lambda y \ ; \ 2 = 2\lambda z \implies 1 = \lambda z\\ \$
which give the following relations for x, y, z:
$x = \frac{3}{4\lambda} \ ; \ y = \frac{3}{4\lambda} \ ; \ z = \frac{1}{\lambda}$
Plugging these values in the constraint:
$2\bigg(\frac{3}{4\lambda}\bigg)^2 + 4\bigg(\frac{3}{4\lambda}\bigg)^2 + \bigg(\frac{1}{\lambda}\bigg)^2 = 70\\ \\ \implies \frac{70}{16\lambda^2} = 70 \implies \lambda^2 = \frac{1}{16} \implies \lambda = \pm\frac{1}{4}$
So, the respective values of x,y,z are:
$x = 3 \ \ or \ \ x = -3 \ ; \ y = 3 \ \ or \ \ y = -3 \ ; \ z = 4 \ \ or \ \ z = -4$
So, we have two critical points: (3,3,4) and (-3,-3,-4). So,
$f(3,3,4) = 35 \ \ and \ \ f(-3,-3,-4) = -35$
So, the maximum value of the function f(x,y,z) is 35 and the minimum value of the function is -35.
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(b)
(i)
$f(x,y) = xye^{x-y} \implies f_x = y(1+x)e^{x-y} \ \ and \ \ f_y = x(1-y)e^{x-y}\$
So, at (5,5)
$f_x = (5)(6)e^0 = 30 \ \ and \ \ f_y = (5)(-4)e^0 = -20$
The unit vector for u(1,2) is:
$u = \frac{1}{\sqrt5}<1,2>$
So, the directional derivative for f(x,y) in the direction of u is:
$<30,-20>.\frac{1}{\sqrt5}<1,2> = \frac{-10}{\sqrt5} = -2\sqrt5$
(ii)
The function increases most rapidly in the direction of its gradient.
$grad \ f(x,y) = <f_x,f_y> = <y(1+x)e^{x-y}, x(1-y)e^{x-y}>$
At point (5,5):
$grad \ f(x,y) = <30,-20> = 10<3,-2>$
So, at (5,5), the function increases most rapidly in the direction <30 , -20>.
(iii)
The maximum rate of change is the absolute value of the gradient at that point, ie.,:
$|grad \ f(x,y)| = |30, -20| = 10\sqrt{13}$
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(c)
$f(x,y) = x^2 + 2y^2 - x^2y + 5 \implies f_x = 2x-2xy \ \ and \ \ f_y = 4y - x^2$
The criticcal points will be when both equations are simultaneously zero.
$f_x = 2x-2xy = 0 \ \ \ and \ \ \ f_y = 4y - x^2 = 0\\ \implies2x(1-y) = 0 \implies x = 0 \ \ or \ \ y = 1\\ \\ \implies y = 0 \ \ or \ \ x = \pm2$
So, critical points are:
$(0,0) ; (2,1) ; (-2,1)$
The double derivatives are:
$f_{xx} = 2-2y \ ; \ f_{yy} = 4 \ ; \ f_{xy} = f_{yx} = -2x$
So, the determinant is calulated as:
$D = f_{xx}f_{yy} - f_{xy}^2 = 4(2-2y) - (-2x)^2 = 8 - 8y -4x^2\\ \\ \implies D_{(0,0)} = 8 \ \ and \ \ f_{xx}|_{(0,0)} = 2$
Since both values are greater than 0, f(x,y) has relative minima at (0,0).
$D_{(2,1)} = -16$
Since D < 0 , f(x,y) has saddle point at (2,1)
$D_{(-2,1)} = -16$
Since D < 0, f(x,y) has saddle point at (-2,1)