Find the equation of the line through P=(8,7) such that the triangle bounded by this line and the axes in the first quadrant has the minimal area.
Answers
The area of the triangle is given by the product of 0.5 and the distances of the x and y intercepts from the origin.
[tex]\mathrm{The \ equation \ of \ the \ line \ is, }\displaystyle \ \underline{ y = 14 - \frac{7}{8} \cdot x}[/tex]Reasons:
The area of the triangle, A = 0.5·x·y
Where;
x and y are the values of the line at the intercepts
The equation of the line is; (y - 7) = m·(x - 8)
At the y-intercept, x = 0, therefore;
(y - 7) = m·(0 - 8) = -8·m
At the y-intercept, y = 7 - 8·m
At the x-intercept, y = 0, therefore;
(0 - 7) = m·(0 - 8) = -8·m
At the x-intercept, -7 = m·x - 8·m
8·m - 7 = m·x
[tex]\displaystyle x = \mathbf{8 - \frac{7}{m}}[/tex]
Therefore;
[tex]\displaystyle A = 0.5 \times \left(7 - 8 \cdot m\right) \times \left(8 - \frac{7}{m} \right) = \mathbf{\frac{-32 \cdot m^2 +56 \cdot m - 24.5}{m}}[/tex]
When the area is minimal, we have;
[tex]\displaystyle \frac{dA}{dm} =0 = \frac{d}{dm} \left(\frac{-32 \cdot m^2 +56 \cdot m - 24.5}{m}\right) = \mathbf{-\frac{32 \cdot m^2 - 24.5}{m^2}}[/tex]
m² × 0 = 24.5 - 32·m²
[tex]\displaystyle m^2 = \frac{24.5}{32} = \frac{49}{64}[/tex]
[tex]\displaystyle m = \sqrt{\frac{49}{64}} = \frac{7}{8}[/tex]
[tex]\displaystyle m = \pm \frac{7}{8}[/tex]
The equation of the line when [tex]\displaystyle m = + \frac{7}{8}[/tex] is therefore;
(y - 7) = m·(x - 8)
[tex]\displaystyle (y - 7) = \mathbf{\frac{7}{8} \cdot (x - 8)}[/tex]
[tex]\displaystyle (y - 7) = \frac{7}{8} \cdot (x - 8) = \frac{7}{8} \cdot x - \frac{7}{8} \times 8 = \frac{7}{8} \cdot x - 7[/tex]
[tex]\displaystyle y = \frac{7}{8} \cdot x - 7 + 7 = \frac{7}{8} \cdot x[/tex]
[tex]\displaystyle y = \mathbf{\frac{7}{8} \cdot x}[/tex]
The x and y intercept of the above line are 0
When [tex]\displaystyle m = - \frac{7}{8}[/tex], we have;
[tex]\displaystyle (y - 7) = -\frac{7}{8} \cdot (x - 8) = -\frac{7}{8} \cdot x + \frac{7}{8} \times 8 = -\frac{7}{8} \cdot x + 7[/tex]
Which gives;
[tex]\displaystyle y = \frac{7}{8} \cdot x + 7 + 7 = \frac{7}{8} \cdot x + 14[/tex]
[tex]\mathrm{The \ equation \ of \ the \ line \ is, }\displaystyle \ y = \mathbf{14 - \frac{7}{8} \cdot x}[/tex]
The equation of the line through P = (8, 7) such that the triangle bounded by the line and the axes in the first quadrant has minimal area is therefore;
[tex]\displaystyle \ \underline{ y = 14 - \frac{7}{8} \cdot x}[/tex]
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