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:
$$ y= -\left( \dfrac{4}{3} \right) x + 8 $$
Best Answer
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 $AB_1 > AB$.
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 $-\frac{b}{a} = -\frac43$.