Help with a high school problem involving some analytical geometry and derivation

[gacicmia]

Through the point (1, 4), draw a line such that the sum of the lengths of the
waits (on the positive parts of the coordinate axes) is the smallest. The result should be x/3+y/6=1

I know It’s simple but I have been looking at it for an hours and I don’t know how to do it

Thanks in advance

 

[stapel]

Through the point (1, 4), draw a line such that the sum of the lengths of the
waits (on the positive parts of the coordinate axes) is the smallest. The result should be x/3+y/6=1

What are "waits"?

 

[Steven G]

What are "waits"?

I heard that you benefited the most from those mind reading classes but I guess after a long absence you forgot most of it. I think it's time that we all meet up and retake that class.

 

[The Highlander]

It seems apparent (to me anyway) that English is not the OP's first language and what s/he is being asked to calculate is the smallest possible sum of the legs of a right triangle whose hypotenuse passes through (1, 4) and the vertex of the right angle lies at the Origin. ie: the sum of the lengths of the orange lines in the diagram below...

​

I suspect the original question might have been (more properly?) translated as: "Through the point (1, 4), draw a line such that the sum of the lengths of the intercepts (on the positive parts of the coordinate axes) is the smallest."; it may be that Google translates "intercept" (or similar term) as "wait" from whatever language the OP speaks?

My approach to the problem (there may be simpler ones) would be as follows...

The lines that pass through the point (1, 4) will have the equation \(\displaystyle y=mx+c\) but (to form the required triangle) \(\displaystyle c > 4\) and \(\displaystyle m < 0\) must be true.

Also, if the point (1, 4) lies on a straight line then \(\displaystyle 4=m\times 1+c\implies \color{red}c=4-m\) and its equation may also be (re)written as: \(\displaystyle y - 4 = m(x-1)\).

To find the intercept on the x-axis we would substitute y = 0 and get...

\(\displaystyle 0-4=m(x-1)\)

\(\displaystyle \implies\frac{-4}{m}=x-1\)

\(\displaystyle \implies x=\frac{m-4}{m}\)

Similarly, to find the intercept on the y-axis we would substitute x = 0 and get...

\(\displaystyle y-4=m(0-1)\)

\(\displaystyle \implies y=-m+4\)

\(\displaystyle \implies y=-(m-4)\)

So, the intercepts on the coordinate axes are: \(\displaystyle \frac{m-4}{m}\) and \(\displaystyle -(m-4)\).

Now, let \(\displaystyle S\) be the sum of the intercepts so...

\(\displaystyle S=\frac{m-4}{m}-(m-4)\)

Then: \(\displaystyle \,\frac{\mathrm{d} S}{\mathrm{d} m}=4m^{-2}-1\)

And, for maximum or minimum values of \(\displaystyle S\), we must have...

\(\displaystyle \frac{\mathrm{d} S}{\mathrm{d} m}=0\)

\(\displaystyle \implies\frac{4}{m^2}-1=0\)

\(\displaystyle \implies\frac{4}{m^2}=1\)

\(\displaystyle \implies m^2=4\)

\(\displaystyle \implies m=\pm\, 2\)

But, since \(\displaystyle m < 0\), then m must be -2 and, since \(\displaystyle \color{red}c=4-m\), then \(\displaystyle c=6\).

Hence; \(\displaystyle y=-2x+6\) (which may be transposed to: \(\displaystyle \frac{x}{3}+\frac{y}{6}=1\)).