Bah calculus calculus & more calculus

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got a simple problem.
the question is find min value for f=x*y+(z^2) with constraints 2*x -y=8, and co-ordinates where it occurs.
so far what i did.
▽f=λ*▽g
F(x,y,z)=f(x,y)-λg(x,y,z)
F=(xy+z^2)-λ(2x-y-8)

y=2λ
x=λ

i got λ=-2

is this right? and is that the only λ value?

how do i even start to find equilibrium solutions & general solution to ((t/2)-2)*sin(y), we weren't taught or shown how to find it involving sin,cos, tan?
for general solution i should take it as separable to find the general solution? but how do i start? sin, cos, tan bah
 
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you could probaby substitute directly into that function & minimise
 
otherwise lagrange multipliers are alway good...
 
mathstruggle said:
got a simple problem.
the question is find min value for f=x*y+(z^2) with constraints 2*x -y=8, and co-ordinates where it occurs.
so far what i did.
▽f=λ*▽g
F(x,y,z)=f(x,y)-λg(x,y,z)
F=(xy+z^2)-λ(2x-y-8)

y=2λ
x=λ

i got λ=-2

is this right? and is that the only λ value?
I think you are trying to use "Lagrange multipliers" as lanedance suggested but you seem to have no idea how to do that.

You write, correctly, that \nabla f= \lambda \nabla g but then "F(x,y,z)=f(x,y)-λg(x,y,z)" and "F=(xy+z^2)-λ(2x-y-8)" which have nothing to do with what you wrote previously.

\nabla f is the vector y\vec{i}+ x\vec{j}+ 2z\vec{k} and \nabla g= 2\vec{i}- \vec{j}.

"\nabla f= \lambda \nabla g" is now
y\vec{i}+ x\vec{j}+ 2z\vec{k}= \lambda 2\vec{i}- \vec{j}

Looking at the individual components of that, y= 2\lambda, x= -\lambda, and z= 0.

Now, you have y= 2\lambda but you have x= \lambda[\itex] rather than x= -\lambda. Perhaps that is just a typo. In any case, they do <b>not</b> give &quot;\lambda= -2 because you have no reason to believe x= -2 or y= -4!.<br /> <br /> Rather, x= -\lambda says that \lambda= -x and so y= 2\lambda= -2x. Putting that into the constraint 2x- y= 8 gives 2x+ 2x= 4x= 8 so x= 2 and y= -4. The solution is (2, -4, 0).<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> how do i even start to find equilibrium solutions &amp; general solution to ((t/2)-2)*sin(y), we weren&#039;t taught or shown how to find it involving sin,cos, tan? <br /> for general solution i should take it as separable to find the general solution? but how do i start? sin, cos, tan bah </div> </div> </blockquote> Is this a completely different problem? Then <b>what</b> is the problem? I associate &quot;equilibrium solutions &amp; general solution&quot; with differential equations but you give no differential equation. Are you talking about dy/dt= ((t/2)- 2)sin(y) or some other problem? Please state the entire problem.<br /> <br /> And a general suggestion: if you want people who are really good at math to help you (hopefully more politely than I did), stop dissing mathematics!
 
HallsofIvy said:
I think you are trying to use "Lagrange multipliers" as lanedance suggested but you seem to have no idea how to do that.

You write, correctly, that \nabla f= \lambda \nabla g but then "F(x,y,z)=f(x,y)-λg(x,y,z)" and "F=(xy+z^2)-λ(2x-y-8)" which have nothing to do with what you wrote previously.

\nabla f is the vector y\vec{i}+ x\vec{j}+ 2z\vec{k} and \nabla g= 2\vec{i}- \vec{j}.

"\nabla f= \lambda \nabla g" is now
y\vec{i}+ x\vec{j}+ 2z\vec{k}= \lambda 2\vec{i}- \vec{j}

Looking at the individual components of that, y= 2\lambda, x= -\lambda, and z= 0.

Now, you have y= 2\lambda but you have x= \lambda[\itex] rather than x= -\lambda. Perhaps that is just a typo. In any case, they do <b>not</b> give &quot;\lambda= -2 because you have no reason to believe x= -2 or y= -4!.<br /> <br /> Rather, x= -\lambda says that \lambda= -x and so y= 2\lambda= -2x. Putting that into the constraint 2x- y= 8 gives 2x+ 2x= 4x= 8 so x= 2 and y= -4. The solution is (2, -4, 0). Is this a completely different problem? Then <b>what</b> is the problem? I associate &quot;equilibrium solutions &amp; general solution&quot; with differential equations but you give no differential equation. Are you talking about dy/dt= ((t/2)- 2)sin(y) or some other problem? Please state the entire problem.<br /> <br /> And a general suggestion: if you want people who are really good at math to help you (hopefully more politely than I did), stop dissing mathematics!
i didn&#039;t diss maths. why would i diss maths if I am majoring in maths? its called sense of humour. last time I am using this forum, its like dictatorship here no freedom.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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