Hey Mark,
I was just told
"To avoid some confusion in this question, I'll point out that you need to have the constraint in the form "g(x, y) - c" in order to get the correct sign for the Lagrange multiplier. In other words, you need to multiply the constraint, as it's given, by minus one."...
Of course. Sorry.
In a situation like this would you state
-\lambda {x}^{-\frac{2}{3}} > 0 for all x > 0 or do you use the value of x we already found x=4 \therefore -\lambda {x}^{-\frac{2}{3}} = 0.125 > 0
Thank you again
Further on this question I need to check the second order conditions.
D(x,y,\lambda) = ({f}_{xx}-\lambda {g}_{xx}){({g}_{y})}^{2} - 2({f}_{xy}-\lambda {g}_{xy}){g}_{x}{g}_{y}+({f}_{yy}-\lambda {g}_{yy}){({g}_{x})}^{2}
{g}_{y}=-1
{g}_{x}=-2{x}^{-\frac{1}{2}}
{g}_{xx}={x}^{-\frac{3}{2}}...
I tried at
y = 0
U(\frac{m}{6}, 0 )
ends at
{(m-8)}^{2} (So on boundary)
Also tried
x = 2
U(2, \frac{m-12}{10})
ends at
{-95m}^{2}-1536m-6080
both outside of the boundary?
U(x(m), y(m)) = U(\frac{40-m}{24}, \frac{m-8}{8}) = \frac{{m}^{2}+16m+64}{96}
Can I choose any point on the constraint? Say
y = 10 - 2x
x(m) = \frac{100-m}{14}
y(m) = \frac{-30-m}{7}
U(x(m), y(m)) = \frac{{3m}^{2}+8m-640}{98}
Original is larger until a certain point?
A consumer spends a positive amount (m) in order to buy (x) units of one good at the price of 6 per unit, and (y) units of a different good at the price of 10 per unit. The consumer chooses (x) and (y) to maximize the utility function
U(x, y) = (x + y)(y + 2).
I need to find the optimal...
A student wishes to minimize the time required to gain a given expected average
grade, 𝑚, in her end-of-semester examinations. Let {t}_{i} be the time spent studying
subject i\in{1,2}.
Suppose that the expected grade functions are {g}_{1}({t}_{1}) = 40+8\sqrt{{t}_{i}} and {g}_{2}({t}_{2}) =...