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  1. K

    Rolle's theorem, to show there's only one root

    Why with the Mean Value Theorem and Rolle's Theorem there is only one root (y=0) if f(a) and f(b) have opposite signs? The Mean Value Theorem says there is at least one root, what is the role of Rolle's theorem in that there is only one root? why does it make this if y'≠0? Rolle's theorem is...
  2. K

    Rolle's theorem, to show there's only one root

    Yes, with the Intermediate Value Theorem: So i even don't need to prove y decreases (by proving y'<0)
  3. K

    Rolle's theorem, to show there's only one root

    $$y'=6x^2 - 6x -12$$ $$\left\{ \begin{array}{l} x_1+x_2=-\frac{b}{a}=1 \\ x_1\cdot x_2=\frac{c}{a}=(-2) \end{array}\right.$$ $$x_1=(-1),~x_2=2$$ And in between y'<0 so y is decreasing, but it doesn't matter since the theorem only demands y'≠0 in the open interval
  4. K

    Rolle's theorem, to show there's only one root

    y'=0 at x=(-2) and x=3, between them y'<0 so y is decreasing So between a=(-1) and b=0 y decreases I don't need y''
  5. K

    Rolle's theorem, to show there's only one root

    Y'=0 at a=(-1) How do i prove y'≠0 on the rest of the open domain <a-b>
  6. K

    Rolle's theorem, to show there's only one root

    Homework Statement Homework Equations Rolle's Theorem: If f(a)=f(b)=0 then there is at least one a<c<b such that f'(c)=0 The Attempt at a Solution $$y=2x^3-3x^2-12x-6~\rightarrow~y'=6x^2-6x-12$$ The function: y': How do i know y' isn't 0 somewhere? if it's continuously descending, so i...
  7. K

    Min max: y=sin x+cos x

    $$y(max)=45^0+2\pi k,~y(min)=225^0+2\pi k$$
  8. K

    Min max: y=sin x+cos x

    Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$y=\sin{x}+\cos{x}~\rightarrow~y'=\cos{x}-\sin{x}$$ $$y'=0:~\rightarrow~\cos{x}=\sin{x}~\rightarrow~x=\frac{\pi}{4}+n\cdot \pi$$ It's not correct
  9. K

    Min max: optimal quantity of medicine

    Yes, that's correct, i need the maximum for R', but why? At the point where R has a maximum, i think, a small change in D makes a big change in R
  10. K

    Min max: optimal quantity of medicine

    $$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2,~~R''=C-2D,~~R''=0:~D=\frac{C}{2}$$ But the greatest change in R for a small change in D is where R has a maximum, hence where R'=0, not where R''=0
  11. K

    Min max: optimal quantity of medicine

    Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$R'=2D\left( \frac{C}{2}-\frac{D}{3} \right)-\frac{1}{3}D^2$$ $$R'=0~\rightarrow~D=C$$
  12. K

    Minimization problem: Economics: quantity to order

    Thank you Ray, you are correct, this is only a minimum problem. I just automatically wrote the heading. Thanks
  13. K

    Minimization problem: Economics: quantity to order

    In the old model: $$A(Q)=\frac{KM}{Q}+cM+\frac{hQ}{2}$$ Where c is the purchase cost of one item. in the new model: $$A(Q)=\frac{K+pQ}{Q/M}+cM+\frac{hQ}{2}=\frac{KM}{Q}+(c+p)M+\frac{hQ}{2}$$ And differentiating gives the same result
  14. K

    Minimization problem: Economics: quantity to order

    Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$Q=\sqrt{\frac{2(K+pQ)}{h}}~\rightarrow~Q=\frac{2}{h}(KM+pM)$$ ##Q'=0~## gives no sense result
  15. K

    Min max: how much of the tax to absorb

    I tried it and it comes right: Genius
  16. K

    Min max: how much of the tax to absorb

    But why: $$\Pi\prime = x(200-0.01x) - 50x + 20,000 - 0.1x$$ i think it should be: $$\Pi\prime = x(200-0.01x) - 50x - 20,000 - 10x$$
  17. K

    Min max: how much of the tax to absorb

    But why ##~\Pi\prime = x(200-0.01x) - 50x + 20,000 - 0.1x~##, i think it should be: $$\Pi\prime = x(200-0.01x) - 50x + 20,000 - 10x$$
  18. K

    Min max: how much of the tax to absorb

    a are cents absorbed by the firm. P is the price presented to the customers, is what they are asked to pay $$\left\{ \begin{array}{l} P-(10-a)=200-0.01x \\ y=(50+a)x+20,000 \\ N=xP-y \end{array}\right.$$ $$N'=0:~0.01x=80-a$$ I don't seem to get rid of a so i will use the result for the original...
  19. K

    Min max: how much of the tax to absorb

    $$\left\{ \begin{array}{l} y=60x+20,000 \\ P=200-0.01x \\ N=(P-a)x-y \end{array} \right.$$ $$N=(140-a)x-0.01x^2-20,000$$ $$N'=0~\rightarrow~140-a=0.02x$$ I have to express x in terms of a, x=x(a)
  20. K

    Min max: how much of the tax to absorb

    ##y = (50 + 10 - a)x + 20,000~## says (10-a) is absorbed by the firm. every item is more expensive to "produce" by (10-a). i shouldn't write ##~N=(P-a)x-y~## since the loss to the firm has already been taken into account in y. So a is passed to the customer. every item is more expensive by a. if...
  21. K

    Min max: how much of the tax to absorb

    a are the cents that are absorbed by the firm. $$N = (P-a)x - y=...=140x-0.01x^2+20,000$$ a disappeared
  22. K

    Min max: how much of the tax to absorb

    The answer to the original question, no' 7, is x=7500. If "all other features are unchanged" then x=7500. $$\left\{ \begin{array}{l} P=200-0.01x-a \\ y=(60-a)x+20,000 \\ x=7500 \end{array} \right.$$ $$N'=0~\rightarrow~x=100(70-a)~\rightarrow~7500=100(70-a)~\rightarrow~a=-5$$
  23. K

    Min max: how much of the tax to absorb

    $$\left\{ \begin{array}{l} P=200-0.01x-a \\ y=(60-a)x+20,000 \end{array} \right.$$ $$N=xP-y=...=2(70-a)x-0.01x^2-20,000$$ $$N'=2(70-a)-0.02x,~~N'=0~\rightarrow~x=100(70-a)$$ It still is with 2 variables. And i see that ##~P=200-0.01x~## represents the ability of the firm to charge the price P if...
  24. K

    Min max: how much of the tax to absorb

    $$\left\{ \begin{array}{l} P=200-0.01x+a \\ y=(60-a)x+20,000 \end{array} \right. $$ $$N=xP-y=...=2(70+a)x-0.01x^2-20,000$$ $$N'=2(70+a)-0.02x,~~N'=0~\rightarrow~x=100(70+a)$$
  25. K

    Min max: how much of the tax to absorb

    Why it's not correct that: ##~y=(50+a)x+20,000~##? it's more expensive to make an item. I don't think you can modify ##~P=200-0.01x~## because it depends only on the customers that are ready to buy x items if the price is P. The formula ##~P=200-0.01x~## doesn't depend at all on the firm, only...
  26. K

    Min max: how much of the tax to absorb

    x is the same, items sold per week, and A is the new coefficient instead of 0.01. A<0.01 since P must be higher now. But maybe P=B-0.01x, i don't know. Or may be even P=B-Ax
  27. K

    Min max: how much of the tax to absorb

    Homework Statement Homework Equations Minimum/Maximum occurs when the first derivative=0 The Attempt at a Solution $$y=20,000+60x,~~P=200-Ax$$ $$N=xP-Y=200x-Ax^2-20,000-60x,~~N'=140-2Ax$$ Two variables
  28. K

    Min max problem

    Thank you Charles, this is the proof. I am not familiar with economics and i felt uncomfortable with this question
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