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So I have H(E,X)=0.01EX and g(x)=0.02[x-0.001x^2) where G=H , so I want to redo the whole thing so everything is function of E, the result should be something like that Y(E)= 10E-0.4E^2 soo..how do I do that? how do I proceed?

Thanks in advance

Maximilian1988
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1 Answers1

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Since $G= H$, you can set the functions equal to one another:

$$0.01EX = 0.02(X-0.001X^2)\iff 0.01 EX = 0.02X(1-0.001X)$$

Now, provided $X \neq 0$, we have $$0.01 E = 0.02(1 - 0.001X)\iff E = 2- 0.002 X\iff \dfrac {2-E}{0.002} = X,\;\;X\neq 0$$

(When $X = 0$, we cannot determine anything about $E$.)

Provided $X\neq 0$, you can express each either G or H as a function of E by substitution.

amWhy
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  • Thanks for your help.what happens though with the 0.001x^2 , why it disappears? – Maximilian1988 Apr 25 '14 at 07:18
  • if its a typo error then the final result is X= (√(2-E)/(0.002)) , what it doesnt add up though is that in the book examples the results are like Y(E)=10E-0.2E2 so where is y here? G and H are y, but in our new equation is x what becomes y? thanks again for your time and help – Maximilian1988 Apr 25 '14 at 08:38
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    It disappears because we factored out an $X$ from $0.02(X-0.001X^2) = 0.02X(1-0.001X)$. Then the $X$ on each side of the equation cancels: $\require{cancel}0.01E\cancel{X} = 0.02\cancel{X}(1 - 0.001X)$, and then simplified to express X as a function of E. – amWhy Apr 25 '14 at 11:15
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    Since $y = G = H$, then $y =\underbrace{0.01EX}{H(X, E)} \iff y = 0.01E\underbrace{\left(\dfrac{2-E}{0.002}\right)}{X} = \dfrac{0.02E - 0.01E^2}{0.002} = 10E - 5E^2.$ – amWhy Apr 25 '14 at 11:20
  • Fantastic, very clear now, sorry I didnt realise of the factorization bit, once again thanks very much for your help, really appreciate it :) – Maximilian1988 Apr 26 '14 at 02:28