Find the equation of the line through the point (4, 7) that cuts off the least area from the first quadrant.

Question

Do step by step.

Answer 1

Let equation of line y = mx + b;; 7 = 4m + b; b = 7 - 4m; x- intercept: 0 = ma + b; a = - b/m = 4 - 7/m;

Area A(m) = ab/2 = (4 - 7/m)(7 - 4m)/2 = 14 - 8m- 49/(2m) + 14 = 28 - 8m - 49/(2m);

A'(m) = - 8 + 49/(2m²) = 0; 16m² = 49; m = ±7/4. Because a > 0 and b > 0 we have to take m = - 7/4

b = 7 - 4(- 7/4) = 14; equation of line y = -7/4x + 14.

This line gives min area because A''(m) = - 49/m³ and A''(- 7/4) = -49/(- 7/4)³ = 64/7 > 0