Find the least value of 3x + 4y if x2y3 = 6.?

Question

plz solve with steps. its answer is 10. but i don't know how .

Answer 1

Using lagrangians multipliers
F(x,y k) = 3x+4y+k(x^2y^3-6)
Fx = 3+2kxy^3=0
Fy= 4 +3kx^2 y^2=0
3kx^2y^2=-4
2kxy^3 =-3
3/2 *x/y= 4/3so x/y= 8/9 and y=9/8x
x^2*(9/8)^3x^3 = 6
x^5 = 6*(8/9)^3 and x = 1.3333 and y = 1.5
4+6=10

Answer 2

The straightforward way to do it is to first find y in terms of x, or

y = (6/x^2)^(1/3)

Plug this in 3x + 4y, and differentiate with respect to x, and set it equal to 0 for min/max, and solve for x, which comes out to x = 4/3. Plug this back into the original equation and get 10.

Addendum: Santmann2002's method of Lagrangian multipliers is the more general way of treating this kind of problem.

Answer 3

x²y³ = 6 y³ = 6/x² y = ³?(6/x²) y = (6^?)/(x^?) y = (6^?)(x^-?) f(x) = 3x + 4(6^?)(x^-?) f'(x) = 3 + 4(6^?)(-?)(x^-5/3) f'(x) = 3 + (-8/3)(6^?)(x^-5/3) 3 + (-8/3)(6^?)(x^-5/3) = 0 (-8/3)(6^?)(x^-5/3) = -3 (8/3)(6^?)(x^-5/3) = 3 (6^?)(x^-5/3) = 3/(8/3) (6^?)(x^-5/3) = 3(3/8) (6^?)(x^-5/3) = 9/8 x^-5/3 = (9/8)/(6^?) x^-5/3 = (9/8)(6^-?) [x^-5/3]?³ = [(9/8)(6^-?)]?³ x? = (9/8)?³(6^-?)?³ x? = 6[(9?³)/(8?³)] x? = 6(8³/9³) x? = 6(512/729) x? = 3072/729 (x?)^(a million/5) = (3072/729)^(a million/5) x = 4/3 y = (6^?)((4/3)^-?) y = a million.5 = 3/2 Least fee: 3x + 4y = 3(4/3) + 4(3/2) = (12/3) + (12/2) = 4 + 6 = 10

Answer 4

Even Ray Charles can see the answer is 10!