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Problem 1:

A surface consists of all points P such that the distance from P to the plane y = 1 is twice the distance from P to the point (0,-1,0). Find an equation for this surface and identify it.

the distance between (0,-1,0) and any point on the surface is:

sqrt(x^2 + (y+1)^2 + z^2)

the formula for the distance between a point on the surface and the plane y = 1 is:

|y-1|/sqrt(1) or just |y-1| (is this correct?)

I came up with this for the equation of the surface:

2*sqrt(x^2 + (y+1)^2 + z^2)=|y-1|

squaring both sides and completing the square for the y term, I ended up with:

Did I do this problem correctly? I wasn't 100% sure which side of the equation to put the 2 on and even if I used the right equations. I may be second guessing myself, but I would really like other peoples opinions.

Problem 2:

Find all critical points and determine whether local maximums or minimums occur at these points for the function f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2

first partials =

fx(x,y) = 6x^2 + y^2 + 10x and fy(x,y) = 2xy + 2y

I sometimes get stumped on simultaneous equations, so I wanted to see if I did this correctly. Critical points occur when both of these equations = 0 simultaneously. I got these points to be f(0,0)=local min, f(-1,-2)=saddle point, f(-1,2)=saddle point, f(-5/3,0)=local max

Was wondering if I did this correctly?

Problem 3:

Find the absolute maximum and absolute minimum values of the function f(x,y)=x^2 - y^2 +xy - 3x on the triangle in the plane whos vertices are (0,0), (3,0), and (3,3)

well, again here I got the partials to be

fx(x,y) = 2x + y - 3 and fy(x,y) = -2y + x

Here I was somewhat confused by the critical points obtained from the line y=x of the triangle. I ended up with 9 critical points to check for absolute max/mins. Here's what I did:

I set the first partials equal to 0 simultaneously and got just 1 point, f(6/5, 3/5)

I took the bottom of the triangle and noticed 0 < or = x < or = 3, and y=0. Puting y=0 into our original function gives f(x,0)=x^2 - 3x. Taking the derivative of this and setting it =0 gave me the point f(3/2,0). Then, looking at the boundaries of this line, you get the points f(0,0) and f(3,0).

In a similar manner for the right side of the triangle, I got the three points f(3,3/2), f(3,0) (which we already have) and f(3,3). So now, I am up to 6 points.

The final line is what was tricky for me. It's equation is y=x, x varies from 0 to 3 inclusive and y also varies from 0 to 3 inclusive. So I thought I had to check the boundary points of f(x,x) and f(y,y).

So for f(x,x), the original function = x^2 - 3x, whose derivative is = 0 when x=3/2. this gave me the point f(3/2,3/2),(I think?). So now(here is where I am not positive in my reasoning) I plugged in the two y values on the boundary and got f(3/2, 0) (already have this) and f(3/2, 3) (which is not in domain (not inside triangle), so we throw it out). Was it a mistake to even consider this one?

for f(y,y), it is similar. we get the original function = y^2 - 3y so y'=0 when y=3/2. this gives the critical point f(3/2,3/2), which we already have. Plugging in the x values on the boundary we get the two points f(0,3/2) (which is not in domain so we throw it out and f(3,3/2) (which we already have).

So, I got the following critical points:

f(6/5,3/5)

f(3/2, 0)

f(0,0)

f(3,0)

f(3,3/2)

f(3,3)

f(3/2,3/2)

f(0,3/2), not in domain

f(3/2,3), not in domain

After checking these, I got f(3/2,0) and f(3/2,3/2) = -9/4 = absolute minimums and f(3,3/2) = 9/4 = the only absolute maximum. Did I do anything wrong here? I feel like I may have found an extra critical point mistakenly in finding the critical points of the third line.

And lastly, problem 4:

A box without a lid is to have a volume of 128,000cm^3. Find the dimensions that minimize the amount of cardboard used.

Well, I used lagrange multipliers, which is how we were supposed to do it.

our function we wish to minimize is:

f(x,y,z) = xy + 2xz + 2yz

our limitation on its domain is:

g(x,y,z) = xyz = 128000

Taking the first partials of each and setting the gradient of f = lamda times the gradient of g, I got the following equations:

y+2z=lamba(yz)

x+2z=lamda(xz)

2x+2y=lamda(xy)

and xyz=128000 (our limitation on the domain)

solving these simultaneous equations, I got x=y=2z. then, solving for one of the variables, I got x=y=cube root of 256000=2z.

I was confident here except for this: why is this a minimum volume and not a maximum volume? did I miss a critical point in solving the simultaneous equations?

Sorry the post is so long, and thanks to anyone for the help. I hope I asked questions adaquately and provided the necessary info.

## Homework Statement

A surface consists of all points P such that the distance from P to the plane y = 1 is twice the distance from P to the point (0,-1,0). Find an equation for this surface and identify it.

## Homework Equations

the distance between (0,-1,0) and any point on the surface is:

sqrt(x^2 + (y+1)^2 + z^2)

the formula for the distance between a point on the surface and the plane y = 1 is:

|y-1|/sqrt(1) or just |y-1| (is this correct?)

## The Attempt at a Solution

I came up with this for the equation of the surface:

2*sqrt(x^2 + (y+1)^2 + z^2)=|y-1|

squaring both sides and completing the square for the y term, I ended up with:

**3(x^2)/4 + [9(y+5/3)^2]/16 + 3(z^2)/4 = 1**, which is the equation of an ellipsoid.Did I do this problem correctly? I wasn't 100% sure which side of the equation to put the 2 on and even if I used the right equations. I may be second guessing myself, but I would really like other peoples opinions.

Problem 2:

## Homework Statement

Find all critical points and determine whether local maximums or minimums occur at these points for the function f(x,y) = 2x^3 + xy^2 + 5x^2 + y^2

## Homework Equations

first partials =

fx(x,y) = 6x^2 + y^2 + 10x and fy(x,y) = 2xy + 2y

## The Attempt at a Solution

I sometimes get stumped on simultaneous equations, so I wanted to see if I did this correctly. Critical points occur when both of these equations = 0 simultaneously. I got these points to be f(0,0)=local min, f(-1,-2)=saddle point, f(-1,2)=saddle point, f(-5/3,0)=local max

Was wondering if I did this correctly?

Problem 3:

## Homework Statement

Find the absolute maximum and absolute minimum values of the function f(x,y)=x^2 - y^2 +xy - 3x on the triangle in the plane whos vertices are (0,0), (3,0), and (3,3)

## Homework Equations

well, again here I got the partials to be

fx(x,y) = 2x + y - 3 and fy(x,y) = -2y + x

## The Attempt at a Solution

Here I was somewhat confused by the critical points obtained from the line y=x of the triangle. I ended up with 9 critical points to check for absolute max/mins. Here's what I did:

I set the first partials equal to 0 simultaneously and got just 1 point, f(6/5, 3/5)

I took the bottom of the triangle and noticed 0 < or = x < or = 3, and y=0. Puting y=0 into our original function gives f(x,0)=x^2 - 3x. Taking the derivative of this and setting it =0 gave me the point f(3/2,0). Then, looking at the boundaries of this line, you get the points f(0,0) and f(3,0).

In a similar manner for the right side of the triangle, I got the three points f(3,3/2), f(3,0) (which we already have) and f(3,3). So now, I am up to 6 points.

The final line is what was tricky for me. It's equation is y=x, x varies from 0 to 3 inclusive and y also varies from 0 to 3 inclusive. So I thought I had to check the boundary points of f(x,x) and f(y,y).

So for f(x,x), the original function = x^2 - 3x, whose derivative is = 0 when x=3/2. this gave me the point f(3/2,3/2),(I think?). So now(here is where I am not positive in my reasoning) I plugged in the two y values on the boundary and got f(3/2, 0) (already have this) and f(3/2, 3) (which is not in domain (not inside triangle), so we throw it out). Was it a mistake to even consider this one?

for f(y,y), it is similar. we get the original function = y^2 - 3y so y'=0 when y=3/2. this gives the critical point f(3/2,3/2), which we already have. Plugging in the x values on the boundary we get the two points f(0,3/2) (which is not in domain so we throw it out and f(3,3/2) (which we already have).

So, I got the following critical points:

f(6/5,3/5)

f(3/2, 0)

f(0,0)

f(3,0)

f(3,3/2)

f(3,3)

f(3/2,3/2)

f(0,3/2), not in domain

f(3/2,3), not in domain

After checking these, I got f(3/2,0) and f(3/2,3/2) = -9/4 = absolute minimums and f(3,3/2) = 9/4 = the only absolute maximum. Did I do anything wrong here? I feel like I may have found an extra critical point mistakenly in finding the critical points of the third line.

And lastly, problem 4:

## Homework Statement

A box without a lid is to have a volume of 128,000cm^3. Find the dimensions that minimize the amount of cardboard used.

## Homework Equations

Well, I used lagrange multipliers, which is how we were supposed to do it.

## The Attempt at a Solution

our function we wish to minimize is:

f(x,y,z) = xy + 2xz + 2yz

our limitation on its domain is:

g(x,y,z) = xyz = 128000

Taking the first partials of each and setting the gradient of f = lamda times the gradient of g, I got the following equations:

y+2z=lamba(yz)

x+2z=lamda(xz)

2x+2y=lamda(xy)

and xyz=128000 (our limitation on the domain)

solving these simultaneous equations, I got x=y=2z. then, solving for one of the variables, I got x=y=cube root of 256000=2z.

I was confident here except for this: why is this a minimum volume and not a maximum volume? did I miss a critical point in solving the simultaneous equations?

Sorry the post is so long, and thanks to anyone for the help. I hope I asked questions adaquately and provided the necessary info.

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