Recent content by Knore88

When we get to finding y: $$20-4\sqrt{x}-y-m=0$$ $$20-8-y-m=0$$ $$12-y-m=0$$ $$y=12-m$$ or do we assume $$\sqrt{4}=-2$$?

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. $$D(x,y,\lambda) = ({f}_{xx}-\lambda {g}_{xx}){({g}_{y})}^{2} - 2({f}_{xy}-\lambda {g}_{xy}){g}_{x}{g}_{y}+({f}_{yy}-\lambda {g}_{yy}){({g}_{x})}^{2}$$ $${g}_{y}=-1$$ $${g}_{x}=-2{x}^{-\frac{1}{2}}$$...

Thank you for the help!

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...

Substituting for m = 70 we have, T = 46 $${t}_{1}$$ = 4 $${t}_{2}$$ = 42 T(4, 42) = 46 for an m of 70?

No problem. T(25, m) = m + 25 (this is larger) But why do we choose this point?

Plugging x = 4 into our constraint we get m - 20 - 4$$\sqrt{4}$$ - y = 0 y = m - 28, and if m = 70, then y = 42?

Does it imply that $$\lambda$$ = -1 and/or x = 4 ?

$${T}_{x}$$(x,y) = 1 $${T}_{y}$$(x,y) = 1 $${g}_{x}$$(x,y) = -$$\frac{2}{\sqrt{x}}$$ $${g}_{y}$$(x,y) = -1 $${T}_{x}$$ = $$\lambda$$*$${g}_{x}$$: 1 = -$$\frac{2}{\sqrt{x}}$$$$\lambda$$ $${T}_{y}$$ = $$\lambda$$*$${g}_{y}$$: 1 = -$$\lambda$$ How does this look?

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...