Probem50+Sophia

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Problem 50 Find an equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

I started off by writing the equation of the line, filling in the information I know y-5 = m (x-3) Because it has to cut off the least area, I know the slope of the line must be negative, so if I look at the area cut off, it is the shape of a right triangle, with sides y¹ and x¹, corresponding to the height and base of the triangle. Because of this, we can use the equation A = 1/2y¹x¹, and solve for those two variables. Because y¹/x¹ is the rise over run, or the slope, solving for these will allow us to find the equation of the line. Simplifying the equation that must be minimized, I subsituted in y¹/x¹ for m. Also, we know that (0,y¹) and (x¹, 0) are points on the line, where the line intersects the axes. By plugging in the point (0,y¹), and solving for x¹ we find that: y¹ - 5 = y¹/x¹ (0 -3) y¹ -5 = -3 (y¹/x¹) y¹ -5 / -3 = y¹/x¹ x¹ = -3y¹ / y¹ -5

if this is plugged into the formula for the area, we find that A = 1/2 (-3y¹ / y¹ -5) y¹ A = 1/2 (-3y¹^2/ y¹-5) You then take the derivative of this equation A' = 1/2 (y¹-5)(-6y¹) - (-3y¹^2) / (y¹ -5)^2 = 1/2 (-6y¹^2 + 30 y¹ + 3 y¹^2) / (y¹ -5)^2 Then, set it equal to 0 0=1/2 (-3y¹^2 + 30y¹)/(y¹-5)2 0 = 1/2 (-3y¹^2 + 30y¹) = -3/2 (-3y¹^2 + 15 y¹) = y¹(-3/2y¹ + 15) 0=y¹, 3/2y¹ = 15 y¹ = 10 Because y¹ = 0 leaves us with x¹= 0, it doesnt work as a solution, because then the line wouldnt cut out a definite part of the quadrant So, we plug in y¹ = 10 x¹ = -3(10)/ (10-5) x¹ = -30/5 x¹ = -6

Plugging the y¹ and x¹ into the original equation, we find that y-5 = 10/-6 (x-3) So, y-5 = -5/3(x-3) is the equation of the line that cuts out the smallest piece of the first quadrant