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3 years ago

the total amount of flour in a bakery after receiving new stock equal to 3/10 of its current stock (x)

Mathematics

1 answer:

Ivenika [448]3 years ago

3 0

If you're look for an equation to find how much flour there would be :

y would be the flour in the bakery after a certain number of stocks based on x, the number of times the flour has been stocked. The equation would require the initial amount of flour, though, which you have not provided.

the constant 1.3 will be used to show that the stock is increasing by 30% each time while including the initial amount of flour.

y= (initial amount of flour)(1.3)^x

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Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

<em>Hence the limit of the function is 3/2.</em>

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