Determine the equation of the line passing by $(3,4)$ which minimizes the area of the triangle determined by its intersection with the coordinate axis

Question

Below is a problem that I did. I do not have faith in my answer. Is my answer right?

Problem:

Consider the equation of a line going through the point $(3,4)$ with a negative slop. It will form a triangle with the y-axis and the x-axis. Find the equation of the that goes through this point and minimizes the area of the triangle.

Answer:

We see a triangle formed by the line and the two axis. The general form of the equation of a line is: $$ y = mx + b$$ In our case, we know that $m$ will be negative. Three points define a triangle. One of those points is $(0,0)$. Another point is $(b,0)$. Now we need to find the point where the line crosses the x-axis. We have: \begin{align*} mx + b &= 0 \\ mx &= -b \\ x &= -\dfrac{b}{m} \end{align*} Hence the third point is $\left( -\dfrac{b}{m}, 0 \right) $. Hence the length of the base is $ \dfrac{b}{x} $. The height of the triangle is $b$. Let $A$ denote the area of the triangle. \begin{align*} A &= \left( \dfrac{1}{2} \right) \left( -\dfrac{b}{m} \right) b \\ A &= - \dfrac{b^2}{m} \\ 4 &= 3m + b \\ b &= 4 - 3m \\ A &= - \dfrac{(4-3m)^2}{m} = - \dfrac{(3m-4)^2}{m} \\ A &= \dfrac{ -9m^2 + 24m - 16}{m} \\ A &= -9m - 24 - 16m^{-1} \\ \end{align*} Now we differentiate $A$ to find a minimal value for $m$. \begin{align*} A' &= -9 + 16m^{-2} \\ -9 + 16m^{-2} &= 0 \\ \dfrac{16}{m^2} &= 9 \\ \dfrac{4}{m} &= -3 \\ m &= -\dfrac{3}{4} \\ b &= 4 - 3\left( -\dfrac{3}{4} \right) = 4 + \dfrac{9}{4} \\ b &= \dfrac{25}{4} \end{align*} Hence the answer is: $$ y= -\left( \dfrac{3}{4} \right) x + \dfrac{25}{4} $$

Here is a corrected / updated solution. However, the answer did not change. Is my revised solution correct?

Now, I see a triangle formed by the line and the two axis. The general form of the equation of a line is: $$ y = mx + b$$ In our case, we know that $m$ will be negative. Three points define a triangle. One of those points is $(0,0)$. Another point is $(b,0)$. Now we need to find the point where the line crosses the x-axis. We have: \begin{align*} mx + b &= 0 \\ mx &= -b \\ x &= -\dfrac{b}{m} \end{align*} Hence the third point is $\left( -\dfrac{b}{m}, 0 \right) $. Hence the length of the base is $ \dfrac{b}{x} $. The height of the triangle is $b$. Let $A$ denote the area of the triangle. \begin{align*} A &= \left( \dfrac{1}{2} \right) \left( -\dfrac{b}{m} \right) b \\ A &= - \dfrac{b^2}{2m} \\ 4 &= 3m + b \\ b &= 4 - 3m \\ A &= - \dfrac{(4-3m)^2}{2m} = - \dfrac{(3m-4)^2}{2m} \\ A &= \dfrac{ -9m^2 + 24m - 16}{2m} \\ A &= \dfrac{-9m - 24 - 16m^{-1}}{2} \\ \end{align*} Now we differentiate $A$ to find a minimal value for $m$. \begin{align*} A' &= \dfrac{-9 + 16m^{-2}}{2} \\ \dfrac{-9 + 16m^{-2}}{2} &= 0 \\ -9 + 16m^{-2} &= 0 \\ \dfrac{16}{m^2} &= 9 \\ \dfrac{4}{m} &= -3 \\ m &= -\dfrac{3}{4} \\ b &= 4 - 3\left( -\dfrac{3}{4} \right) = 4 + \dfrac{9}{4} \\ b &= \dfrac{25}{4} \end{align*} Hence the answer is: $$ y= -\left( \dfrac{3}{4} \right) x + \dfrac{25}{4} $$

Here is an updated solution. Is is it now correct?

Now, I see a triangle formed by the line and the two axis. The general form of the equation of a line is: $$ y = mx + b$$ In our case, we know that $m$ will be negative. Three points define a triangle. One of those points is $(0,0)$. Another point is $(b,0)$. Now we need to find the point where the line crosses the x-axis. We have: \begin{align*} mx + b &= 0 \\ mx &= -b \\ x &= -\dfrac{b}{m} \end{align*} Hence the third point is $\left( -\dfrac{b}{m}, 0 \right) $. Hence the length of the base is $ \dfrac{b}{x} $. The height of the triangle is $b$. Let $A$ denote the area of the triangle. \begin{align*} A &= \left( \dfrac{1}{2} \right) \left( -\dfrac{b}{m} \right) b \\ A &= - \dfrac{b^2}{2m} \\ 4 &= 3m + b \\ b &= 4 - 3m \\ A &= - \dfrac{(4-3m)^2}{2m} = - \dfrac{(3m-4)^2}{2m} \\ A &= \dfrac{ -9m^2 + 24m - 16}{2m} \\ A &= \dfrac{-9m - 24 - 16m^{-1}}{2} \\ \end{align*} Now we differentiate $A$ to find a minimal value for $m$. \begin{align*} A' &= \dfrac{-9 + 16m^{-2}}{2} \\ \dfrac{-9 + 16m^{-2}}{2} &= 0 \\ -9 + 16m^{-2} &= 0 \\ \dfrac{16}{m^{-2}} &= 9 \\ 16 = 9m^2 \\ -4 &= 3m \\ m &= \dfrac{-4}{3} \\ b &= 4 - 3 \left( -\dfrac{4}{3} \right) = 4 + 4 \\ b &= 8 \\ \end{align*} Hence the answer is: $$ y= -\left( \dfrac{4}{3} \right) x + 8 $$

Answer 1

Two (small) mistakes:

(1) You dropped the factor of $\tfrac12$ in your area formula, so you're actually minimizing the area of the rectangle with this diagonal. This shouldn't affect your final answer though, as the minimum will occur at the same slope!

(2) When solving for $m$, you flipped the fraction. The critical point is actually at $m = -\tfrac43$. This affects the final equation of the line, of course. Full solution is hidden below.

The line that minimizes the area of the triangle is $$y = -\tfrac43 x + 8.$$ If you write the equation in standard form, it looks like $$4x + 3y = 24$$ in which you can easily recognize the coordinates $(a, b) = (3, 4)$ and the area $2ab = 24$. Hence, the solution to the general problem with point $(a, b)$ is the line $$bx + ay = 2ab$$ with area $A = 2ab$.

Answer 2

Notice that $y = mx + b = 0$ implies that $x = -b/m$. Notice as well that $x = 0$ implies that $y = b$.

Since the point $(3,4)$ belongs to such line, we conclude that $4 = 3m + b$.

Consequently, the area that we are interested in is given by: \begin{align*} A & = \frac{1}{2}\left(-\frac{b}{m}\right)b = -\frac{b^{2}}{2m} = -\frac{(4 - 3m)^{2}}{2m} = -\frac{9m^{2} - 24m + 16}{2m} \end{align*}

which is (almost) the function that you have obtained: you forgot to divide by $2$.

Hopefully this helps!

Answer 3

If the line makes an angle of $\theta$ with the negative direction of the $x$ axis, then its unit normal vector is $(\sin \theta, \cos \theta)$, and therefore its equation is

$ \sin \theta (x - 3) + \cos \theta (y - 4) = 0$

The $x$ intercept is $ \dfrac{ 3 \sin \theta + 4 \cos \theta }{ \sin \theta } $

The $y$ intercept is $ \dfrac{ 3 \sin \theta + 4 \cos \theta } {\cos \theta } $

So the area of the triangle is

$ A = \dfrac{1}{2} \dfrac{ (3 \sin \theta + 4 \cos \theta)^2 }{ \sin \theta \cos \theta } $

And this simplifies to

$ A = \dfrac{ (12.5 + 3.5 \cos(2 \theta) + 12 \sin(2 \theta)) }{ \sin(2 \theta )} $

Differentiating, and equating the numerator to zero, gives us,

$ - 7 \sin^2 (2 \theta) + 24 \cos(2 \theta) \sin(2 \theta) - 25 \cos(2 \theta) - 7 \cos^2(2 \theta) - 24 \cos(2\theta) \sin(2\theta) = 0$

and this simplifies to,

$ -7 - 25 \cos(2 \theta) = 0 $

Hence,

$ \cos(2 \theta) = -\dfrac{7}{25} $

$ 2 \cos^2(\theta) - 1 = \dfrac{7}{25} $

$ \cos^2 (\theta) = \dfrac{9}{25} $

From which $\cos \theta = \dfrac{3}{5} , \sin \theta = \dfrac{4}{5} $

Therefore, the equation of the line is

$ \dfrac{4}{5} (x - 3) + \dfrac{3}{5} (y - 4) = 0 $

which simplifies to

$ 4 (x - 3) + 3 (y - 4) = 0 $

i.e.

$ 4 x + 3 y - 24 = 0 $

So that

$ y = - \dfrac{4}{3} x + 8 $

Answer 4

Let the side on $y$-axis be $4+b$ and on $x$-axis be $3+a$. By similarity $\frac b3=\frac 4a$ hence $ab=12$.

The area of the triangle is $$A=\frac12 (4+b)(3+a)=12+2a+\frac{18}a=24+2\frac{(a-3)^2}{a}\geq24$$ Equality occurs when $a=3$ and hence $b=4$. The line is passing through $(0,8)$ and $(6,0)$. So it is $y=-\frac43 x+8$.

Answer 5

Given a rectangle $ARPQ$ with sides $\overline{AR} = a$ and $\overline{QA} = b$, we claim that, among the right-angled triangles circumscribed to $ARPQ$ (see figure below), the one with the smaller area is $ABC$ with $\overline{AB} = 2a$ and $\overline{AC} = 2b$.

Consider another right-angled triangle $AB_1C_1$, and suppose first $AB_1 < AB$, as in the figure above, where $BH$ and $CK$ are perpendicular to $B_1C_1$. Denoting with $[\cdots]$ the area of the triangle in parenthesis, we have \begin{eqnarray} [AB_1C_1] &=& [ABC] + [PCK] + [CKC_1] - ([PBH]-[BHB_1])=\\ &=& [ABC] + [CKC_1] + [BHB_1] > [ABC], \end{eqnarray} where we used the fact that $PCK \cong PBH$ (ASA criterion).

I leave you as an exercise to work out in a similar manner the case where $AB_1 > AB$.

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So the triangle with minimum area has area $[ABC] = 2[ARPQ] = 2ab$. In your case $A$ is the origin of the axes, and $P$ is the point having coordinates $(3,4)$. So the mimimum area is $24$ and the slope of the straight line is $-\frac{b}{a} = -\frac43$.