Question

Suppose the wholesale price of a certain brand of medium-sized eggs (p) in dollars per carton is related to the weekly supply (x) in thousands of cartons by the equation

625p^2-x^2 =100

If 25,000 cartons of eggs are available at the beginning of a certain week and the price is falling at the rate of 2c/carton/week, at what rate is the weekly supply falling?

Answer 1

Hi

Answer 2

So if you set x to equal 25 you can solve for p

Answer 3

Hi

Answer 4

How old,r u?

Answer 5

I got p=$1.40

Answer 6

?

Answer 7

Solving for x'

\[625(2)p-2xx'=0 \]

\[\frac{ 1250p }{ 2x}=x'\]

\[\frac{ 1250(1.4) }{ 2(0.2)}=5\]

So is the weekly supply falling at 5,000 a week?

Answer 8

Whops I meant to put \[\frac{ 1250(0.2) }{ 2(25) }= 5\]

Answer 9

hmm where did you get 0.2 for p?
and why did you substitute x' by 5

Answer 10

2c per carton per week is the rate of change for p
meaning p'=2c=0.02$

Answer 11

P changes by unit of cartoon and with weeks
x changes only with week

Answer 12

you need to redo your implicit differentiation

Answer 13

you need to rewrite the problem:
p changes with how many cartoons we would write is as p(x)
x changes by weeks x(w)

Answer 14

the implicit equation is like this
\(\huge \rm \frac{d}{dt}\left(x\right) =\frac{625p \frac{d}{dt}\left(p\right)}{x}\)

Answer 15

and you have \(\huge \rm p=\sqrt{\frac{100+x^2}{625}}\)

Answer 16

Check this later @OdinMW

Answer 17

Hi @xapproachesinfinity thanks for your help!
So I changed the equation to \[\frac{ 625p^2 }{ x }=x\]

and I should solve for d/dt and not d/dx?

Answer 18

\[\frac{ (x)1250p \frac{ dx }{ dt } - \frac{ dx }{ dt }(625p^2)}{ x^2 }=\frac{ dx }{ dt }\]

Answer 19

@freckles hey what do you think?

Answer 20

That won't work! Use the equation i give you

Answer 21

@jim_thompson5910 check this^_^

Answer 22

This is what I get.

x is the supply of eggs in thousands of cartons
p is the price in dollars
t is the time in weeks


"the price is falling at the rate of 2c/carton/week" so dp/dt = -0.02


"If 25,000 cartons of eggs are available at the beginning of a certain week", then we know x = 25. Use this to find the price p

\[\Large 625p^2-x^2 =100\]
\[\Large 625p^2-25^2 =100\]
\[\Large 625p^2-625 =100\]
\[\Large 625p^2 =100+625\]
\[\Large 625p^2 =725\]
\[\Large p^2 = \frac{725}{625}\]
\[\Large p^2 = \frac{29}{25}\]
\[\Large p = \sqrt{\frac{29}{25}}\]
\[\Large p = \frac{\sqrt{29}}{\sqrt{25}}\]
\[\Large p = \frac{\sqrt{29}}{5}\]
\[\Large p \approx 1.07703\]
\[\Large p \approx 1.08\]
The price of the carton of eggs is approximately $1.08
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Implicitly differentiate both sides with respect to t
Plug in p = 1.08, dp/dt = -0.02 and x = 25
Then isolate dx/dt
\[\Large 625p^2-x^2 =100\]
\[\Large \frac{d}{dt}[625p^2-x^2] = \frac{d}{dt}[100]\]
\[\Large \frac{d}{dt}[625p^2]-\frac{d}{dt}[x^2] = 0\]
\[\Large 625*2*p*\frac{dp}{dt}-2x*\frac{dx}{dt} = 0\]
\[\Large 625*2*(1.08)*(-0.02)-2(25)*\frac{dx}{dt} \approx 0\]
\[\Large -27-50\frac{dx}{dt} \approx 0\]
\[\Large -50\frac{dx}{dt} \approx 27\]
\[\Large \frac{dx}{dt} \approx \frac{27}{-50}\]
\[\Large \frac{dx}{dt} \approx -0.54\]
dx/dt is negative to reflect a falling supply

Answer 23

since x is the number of cartons of eggs in thousands and dx/dt is approximately -0.54, this means the supply shrinks by 0.54*1000 = 540 cartons in this time period

Answer 24

ah yes that's the same thing that i was did
I didn't evaluate! i just wanted someone to check the work^_^