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The problem goes like this-

A chemical process is used to remove impurities from a pulp by bleaching process generating toxic and non-toxic wastes.

The toxic waste generated is 0.05 times the square of the amount of chemical used, while the non-toxic waste generated is 0.1 times the amount of chemical used. The pulp yield is 19.1 times the amount of chemical used.

Determine the amount of chemical to be used to maximize the pulp yield and minimize the waste?

My approach is trying to form single variable functions for the total waste & yield. Then take the ratio of yield and total waste and try to maximize it. The derivative is coming out to be decreasing function with value tending to infinity at the boundry (chemical used=0). I am stuck at this limbo.

Any improvement of this process or new approach would be of great help to me

Ash
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1 Answers1

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Let $f(x)$ be the total pulp produced per $x$ unit of chemicals and $g(x)$ be the total amount of waste produced per $x$ unit of chemicals.

Our goal is to maximise $f(x)$ and minimise $g(x)$. So consider a function $C(x) = f(x)-g(x)$, which will be maximised when $f(x)$ is maximum and $g(x)$ is minimum. Now given $f(x) = 19.1x$ and $g(x) = 0.05x^2 + 0.1x \Rightarrow$ $C(x) = -0.05x^2 + 19x$

I shall give two solutions for your problem, as follows:

Solution 1

Now $C(x)$ is a simple parabola that opens downwards as shown below, and thus its maxima is its vertex. The vertex for a quadratic function $h(x) = ax^2 + bx + c$ is given by the point with coordinates $\left(\frac{-b}{2a}, \frac{-D}{4a}\right)$ where $D = b^2 - 4ac$ is the discriminant. A proof of the formula for the vertex can be found here.

Graph of C(x)

Applying the formula, you will find for what $x$ $C(x)$ is the maximum, and how much the maximum value is (the $y$).

Solution 2

If you were to differentiate $C(x)$ with respect to $x$, you will find a linear decreasing function. We can set this function $C'(x) = 0$ which gives us the inflection point at $x=190$. As the second derivative $C''(x) = -0.1 < 0$, the maxima is at $x=190$ by the second-order derivative test. The proof for the second-order derivative test can be found here

Kraken
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  • I'm not convinced. The maxima at $x=190$, are these $g$ or $kg$ or $t$? In case of $g$ -- how to scale-up? – m-stgt Jun 17 '24 at 09:10
  • I can't answer that, as the OP has not provided any information about the units. This is also likely to be a simple calc 1 problem OP encountered on a test, homework, or something else, so we need not worry about such trivial things. And what do you mean by 'scaling up'? – Kraken Jun 17 '24 at 10:50
  • Thanks for the revert. This seems completely logical. As far as units are concerned that is trivial. – Ash Jun 17 '24 at 11:47
  • No problem, hopefully you learnt something new! – Kraken Jun 17 '24 at 11:59