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# Differentiation

Given a curve $$128x^2 - 16x^2y + 1 = 0$$, how do you find the minimum value of $$(x+y)$$?

Is it possible to do this?

$128x^2-16x^2y+1=0\Rightarrow y = 8 + \frac{1}{16x^2}$

$\therefore x+y=8+x+\frac{1}{16x^2}$

$\frac{d(x+y)}{dx}=1-\frac{1}{8x^3}$

Subsequently, equate the above derivative to 0 to find the stationary value of x and its corresponding value of y to find the minimum value of x + y.

Is this legit?

Note by Ho Wei Haw
4 years, 4 months ago

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Actually, what you are doing is not so wrong. There is no minimum value of $$x+y$$ for your function as defined, but there is if you restrict attention to the region $$x>0$$. You just need to change variable. If we put $$z=x+y$$, then $$z = 8 + x + \tfrac{1}{16x^2}$$, and it is clear that $$z$$ does have a minimum of $$8\tfrac34$$ over the domain $$x>0$$, which it achieves at $$x=\tfrac12$$.

alt text

If $$\hat{x}>0$$ and $$\hat{y}$$ are such that $$128\hat{x}^2 - 16\hat{x}^2\hat{y} + 1 = 0$$, then $$\hat{z}=\hat{x}+\hat{y}$$ is such that $$\hat{z} = 8 + \hat{x} + \tfrac{1}{16\hat{x}^2}$$, and hence $$\hat{z} \ge 8\tfrac34$$. But this tells us that $$\hat{x} + \hat{y} \ge 8\tfrac34$$, and hence $$8\tfrac34$$ is a minimum value of $$x+y$$ on this curve. By moving from $$(x,y)$$ to $$(x,z)$$ you are simply changing the coordinate system, and considering a pair of skew axes. It is often OK to do this.

- 4 years, 4 months ago

Yes! This is exactly how I did to solve the question. However, whatever George said made sense too...

- 4 years, 4 months ago

George was right that there was no global minimum for $$x+y$$. He was also right in finding the value of $$x$$ where $$x+y=0$$. What you wanted to do, and did, was find a good way of minimizing $$x+y$$ for positive $$x$$.

- 4 years, 4 months ago

Thanks! :) Greatly appreciate your help :D

- 4 years, 4 months ago

The problem as stated makes no sense. Let $$y(x) = 8 + (16x^2)^{-1}$$, now note that as $$x \to -\infty$$, we have $$y = 8$$. But as $$x \to -\infty$$, $$x + y$$ goes to $$-\infty$$, so there is no absolute minimum.

Perhaps something like $$| x + y |$$ was meant? Either way, this process is not "legit". You cannot take the derivative with respect to the sum of two variables (your notation is also somewhat of a problem, $$\frac{d(x+y)}{dx}$$ means that the derivative of x+y wrt x). You'll need to use Lagrange Multipliers or use a different argument

If you meant that it should be $$|x+y|$$, then let:

$j = \frac{1}{48} \left(4 \sqrt[3]{2 \left(16411+3 \sqrt{98385}\right)}+\sqrt[3]{2100608-384 \sqrt{98385}}+128\right)$

The solution is $$(-j, j)$$. Clearly, the lowest possible thing that $$|x+y|$$ can be is zero, so we want to find the point where $$x = -y$$ (which will make this absolute value 0). Therefore, we need the solution of $$128y^2 + 16y^3 + 1 = 0$$ (which is $$y = j$$). You can use the discriminant of the polynomial to show that $$j$$ exists without calculating it.

- 4 years, 4 months ago

Thanks! Unfortunately, this question came out in my mathematics test and the question clearly asked for the minimum value of (x + y).

However, differentiation is the only process we are supposed to know for our syllabus to locate the maxima and minima of a function.

- 4 years, 4 months ago

its easy...find the value of y from the expression and use to calculate the value of (x+y). in terms of x. Differentiate the expression.As you know at minima point dy/dx=0 and d2y/dx2>0; hence use it to get the value of x. x will come 1/2. put the value in the expression of (x+y) to get the desired result...

- 4 years, 4 months ago