Math Help - 2 weird problems involving implicit differentiation

1. 2 weird problems involving implicit differentiation

The first problem:

(x^2+y^2+5x)^2 = 25/2 (x^2+y^2) , find the equation of the tangent line to P(5,5)

1. y+ 1/3 x = 20/3
2. y = 1/3 x + 5/3
3. y+ 3x + 20/3 = 0
4. y + 1/3 x =5/3
5. y = 3x + 20/3

I always derive the eqn and plug in (5,5) but i always get a slope of -13/15 or something like that, and theres too much work to list out here, so i just really dont know

Problem 2:

A company manufacturing lawn mowers find that its weekly cost and demand equations are

C(x) = 40 + 24x, p(x) = 40 − x/4

when it sells x lawn-mowers per week. If the company can sell all the lawn-mowers it manufactures, determine the maximum profit from the sale of these lawn-mowers.

1. maximum profit = $211 2. maximum profit =$221
3. maximum profit = $201 4. maximum profit =$216
5. maximum profit = $206 I honestly have no idea where to start on this one Thanks for anyone who helps! Anthony 2. Originally Posted by cubejunkies The first problem: (x^2+y^2+5x)^2 = 25/2 (x^2+y^2) , find the equation of the tangent line to P(5,5) 1. y+ 1/3 x = 20/3 2. y = 1/3 x + 5/3 3. y+ 3x + 20/3 = 0 4. y + 1/3 x =5/3 5. y = 3x + 20/3 I always derive the eqn and plug in (5,5) but i always get a slope of -13/15 or something like that, and theres too much work to list out here, so i just really dont know Problem 2: A company manufacturing lawn mowers find that its weekly cost and demand equations are C(x) = 40 + 24x, p(x) = 40 − x/4 when it sells x lawn-mowers per week. If the company can sell all the lawn-mowers it manufactures, determine the maximum profit from the sale of these lawn-mowers. 1. maximum profit =$211
2. maximum profit = $221 3. maximum profit =$201
4. maximum profit = $216 5. maximum profit =$206

I honestly have no idea where to start on this one

Thanks for anyone who helps!
Anthony
For(1),

Actually, there isn't really too much work. Just differentiate implicitly as usual, you do not have to rearrange to make dy/dx the subject but plug in the point straightaway instead.

3. I did, but I keep getting -13/15 as the slope of the tangent line, so, idk, i just need a walkthrough or something :/

4. Originally Posted by cubejunkies
I did, but I keep getting -13/15 as the slope of the tangent line, so, idk, i just need a walkthrough or something :/
I would like to walk through but that wouldn't be possible without seeing your working.

5. (x^2+y^2-5x)(x^2+y^2-5x) = 25/2 x^2 + 25/2 y^2
x^4 - 10x^3 -10xy^2 +2(y^2)(x^2)+ y^4 +25x = 25/2 x^2 + 25/2 y^2
4x^3 - 30x^2 -20xy(dy/dx) +4(x^2)y(dy/dx)+ 4xy^2 +4y^3 (dy/dx) +25x -25y = 0

(-4x^3+30x^2+10y-4(y^2)x-25x)/(-20xy+4(x^2)y+4y^3-25y) = dy/dx

and substituting in 5 for x and 5 for y yielded

-325/375 which simplifies to -13/15

this is the condensed version of my work, i was very tedious in my expansion and omitted steps here

6. Originally Posted by cubejunkies
(x^2+y^2-5x)(x^2+y^2-5x) = 25/2 x^2 + 25/2 y^2
x^4 - 10x^3 -10xy^2 +2(y^2)(x^2)+ y^4 +25x = 25/2 x^2 + 25/2 y^2
4x^3 - 30x^2 -20xy(dy/dx) +4(x^2)y(dy/dx)+ 4xy^2 +4y^3 (dy/dx) +25x -25y = 0

(-4x^3+30x^2+10y-4(y^2)x-25x)/(-20xy+4(x^2)y+4y^3-25y) = dy/dx

and substituting in 5 for x and 5 for y yielded

-325/375 which simplifies to -13/15

this is the condensed version of my work, i was very tedious in my expansion and omitted steps here
There is no need to expand.

$
(x^2+y^2+5x)^2=\frac{25}{2}(x^2+y^2)$

$2(x^2+y^2+5x)(2x+2y\frac{dy}{dx}+5)=\frac{25}{2}(2 x+2y\frac{dy}{dx})$

7. and then what? i plugged in (5,5) and got a slope of -65/11 ? I'm completely lost

8. Originally Posted by cubejunkies
and then what? i plugged in (5,5) and got a slope of -65/11 ? I'm completely lost
How did you get that? I got -17/11 .

9. well, that still isnt one of the slopes of the potential lines for answers :/

10. Originally Posted by cubejunkies
well, that still isnt one of the slopes of the potential lines for answers :/
That is because you made a typo in the equation.

$
(x^2+y^2+5x)^2=\frac{25}{2}(x^2+y^2)$

It should have been (x^2+y^2-5x)^2=\frac{25}{2}(x^2+y^2) (in red)

using this, you would get the correct gradient.