Find the slope and y-intercept of the line through the point (8,6) that cuts off the least area from the first quadrant.

Question

slope = 

y-intercept = 

Answer 1

A = 1/2bh = 1/2x₂y₁

Using similar triangles y₁/6 = x₂/(x₂ - 8) so y₁ = 6x₂/(x₂ - 8) so, subbing this into the area equation:

A(x₂) = 1/2x₂(6x₂/(x₂ - 8))

A(x₂) = 3x₂²/(x₂ - 8))

To minimize the function, take the derivative and set it equal to x=zero.

To take the derivative, we must use the quotient rule.

A'(x₂) = [(3x₂²)'(x₂ - 8) - (3x₂²)(x₂ - 8)'/(x₂ - 8)²

A'(x₂) = [(6x₂)(x₂ - 8) - (3x₂²)(1)'/(x₂ - 8)²

A'(x₂) = (6x₂² - 48x₂ - 3x₂²)/(x₂ - 8)²

A'(x₂) = (3x₂² - 48x₂)/(x₂ - 8)²

Setting this equal to zero gives:

(3x₂² - 48x₂)/(x₂ - 8)² but this can only be equal to zero if the numerator is zero so:

3x₂² - 48x₂ = 0

3x₂(x² - 16) = 0

so x₂ = 0 or x₂ = 16

We can throw out the first solution and go with x₂ = 16

y₂ = 0 so our line goes through (8. 6) and (16, 0) meaning the slope is (0 - 6)/(16 - 8) = -6/8 = -3/4

The y intercept is y1 which we said was y₁ = 6x₂/(x₂ - 8) = 6•16/(16-8) = 12

The slope is -3/4 and the y-intercept is 12