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?

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TopNewestActually, 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 \(x=\tfrac12\).

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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. – Mark Hennings · 3 years, 8 months ago

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– Ho Wei Haw · 3 years, 8 months ago

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

– Mark Hennings · 3 years, 8 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 \(x\).Log in to reply

– Ho Wei Haw · 3 years, 8 months ago

Thanks! :) Greatly appreciate your help :DLog in to reply

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. – George Williams · 3 years, 8 months ago

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However, differentiation is the only process we are supposed to know for our syllabus to locate the maxima and minima of a function. – Ho Wei Haw · 3 years, 8 months ago

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– Sarojit Auddya · 3 years, 8 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...Log in to reply