# 2 weird problems involving implicit differentiation

• Sep 8th 2010, 08:14 PM
cubejunkies
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 • Sep 8th 2010, 08:49 PM mathaddict Quote: 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.
• Sep 8th 2010, 09:00 PM
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 :/
• Sep 8th 2010, 09:23 PM
Quote:

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.
• Sep 8th 2010, 09:32 PM
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
• Sep 8th 2010, 09:46 PM
Quote:

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)$

$2(x^2+y^2+5x)(2x+2y\frac{dy}{dx}+5)=\frac{25}{2}(2 x+2y\frac{dy}{dx})$
• Sep 8th 2010, 09:54 PM
cubejunkies
and then what? i plugged in (5,5) and got a slope of -65/11 ? I'm completely lost
• Sep 8th 2010, 10:01 PM
Quote:

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 .
• Sep 8th 2010, 10:09 PM
cubejunkies
well, that still isnt one of the slopes of the potential lines for answers :/
• Sep 8th 2010, 10:15 PM
Quote:

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.