Find the slope and y-intercept of the line through the point (4,7) that cuts off the least area from the first quadrant.

Question

need slope and y-intercept

Answer 1

a line through that point would be y-7 = m(x-4) where m = the slope

slope = -7/4

just take the y value of the point (4,7) and put it over the x value, slope is rise over run = 7/4 but change the sign to -7/4

y intercept = 14 (2 times 7)

x intercept = 8 (2 times 4)

Area = 1/2(product of intercepts) = (1/2)(14)(8) = 56 = minimum area cut off by the line with slope 7/4 going through (4,7)

equation is

y= -7/4(x) + 14 or multiply to eliminate the fratcion

4y =-7x + 56

7x +4y = 56

another way to see this is find the x and y intercepts in terms of m

set Area = (1/2)(product of those intercepts) and take the derivative, set it equal to zero, then solve for m

the area of a triangle is one half the base times height. the base is the x intercept. the height is the y intercept.

y = m(x-4) + 7

0 = mx -4m + 7

x = 4-7/m = x intercept

y=m(0-4) + 7 = 7-4m = y intercept

Area = (1/2)(product of intercepts)

= (1/2)(7-4m)(4-7/m) = (1/2)(28-16m-49/m+28)

=28 -8m -49/2m

Derivative of A

= A' = -8 +49/2m^2

m^2 = 49/16

m = -7/4

y intercept =7-4m = 7-4(-7/4) = 14