What is the solution of 3x + 5 = 2x – 7? the x-coordinates of the intersection point of the lines y = 3x + 5 and y = 2x – 7 the
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Answer:
The maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)
Step-by-step explanation:
f(x,y) = 3xy, lets find the gradient of f. First lets compute the derivate of f in terms of x, thinking of y like a constant.
In a similar way
Thus,
The restriction is given by g(x,y) = 121, with g(x,y) = 3x²+3y²-5xy. The partial derivates of g are
[ŧex] g_x(x,y) = 6x-5y [/tex]
Thus,
For the Langrange multipliers theorem, we have that for an extreme (x0,y0) with the restriction g(x,y) = 121, we have that for certain λ,
This can be translated into
If we sum the first two expressions, we obtain
Thus, x = -y or λ=3.
If x were -y, then we can replace x for -y in both equations
3y = -11 λ y
-3y = 11 λ y, and therefore
y = 0, or λ = -3/11.
Note that y cant take the value 0 because, since x = -y, we have that x = y = y, and g(x,y) = 0. Therefore, equation 3 wouldnt hold.
Now, lets suppose that λ=3, if that is the case, we can replace in the first 2 equations obtaining
- 3y = 3(6x-5y) = 18x -15y
thus, 18y = 18x
y = x
and also,
- 3x = 3(6y-5x) = 18y-15x
18x = 18y
x = y
Therefore, x = y or x = -y.
If x = -y:
Lets evaluate g in (-y,y) and try to find y
g(-y,y) = 3(-y)² + 3y*2 - 5(-y)y = 11y² = 121
Therefore,
y² = 121/11 = 11
y = √11 or y = -√11
The candidates to extremes are, as a result (√11,-√11), (-√11, √11). In both cases, f(x,y) = 3 √11 (-√11) = -33
If x = y:
g(y,y) = 3y²+3y²-5y² = y² = 121, then y = 11 or y = -11
In both cases f(11,11) = f(-11,-11) = 363.
We conclude that the maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)