How do you minimize and maximize #f(x,y)=x^3-y# constrained to #x-y=4#?

Answer 1

we'll do it first as a problem in single variable calculus.

we have #f(x,y)=x^3-y# but also the constraint #x-y = 4# so we can sub the constraint in to get
#f(x)=x^3-(x-4)#

it's all the usual stuff from here on

#f' = 3x^2 -1# #f'' = 6x#
so #f' = 0 implies x = pm 1/sqrt3#
and #f''(1/sqrt3) >0 # so min
#f''(-1/sqrt3) <0 # so a max

i'll do a Lagrange Multiplier next to compare. the basic premise is that with

#f(x,y)=x^3-y# to optimise
plus the constraint #g(x,y)= x-y = const#
we can say that #nabla f = lambda nabla g#

or

#((3x^2),(-1)) = lambda ((1),(-1))#
#implies lambda = (3x^2)/1 = -1/-1#
So #3x^2 = 1# with same conclusion, stationary points at #x = pm 1/sqrt3# and from the constraint #y = pm 1/sqrt3 - 4#

I'm asking for a second opinion on this next bit.

Because there is no simple way to explore the nature of the turning points, especially with more complex problems, when using the LM approach. You can often play with the physical reality and reason a solution but there is no quick second derivative check, sadly, that i am aware of.

I am just wondering if the Hessian for let's say #L(xy) = x^3-y + lambda(x-y) = const#

would be of any use here.

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Answer 2

To minimize and maximize ( f(x, y) = x^3 - y ) constrained to ( x - y = 4 ), follow these steps:

  1. Solve the constraint equation ( x - y = 4 ) for ( y ) to get ( y = x - 4 ).
  2. Substitute ( y = x - 4 ) into the objective function ( f(x, y) = x^3 - y ) to get the function in terms of ( x ) only: ( g(x) = x^3 - (x - 4) ).
  3. Find the critical points of ( g(x) ) by setting its derivative equal to zero and solving for ( x ).
  4. Determine the corresponding ( y )-values for each critical point using the constraint equation.
  5. Evaluate the objective function ( f(x, y) ) at each critical point to find the maximum and minimum values.

Alternatively, you can use Lagrange multipliers to solve this constrained optimization problem.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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