Hi i need help solving the following

Question

A line is drawn through the point P(3, 4) so that it intersects the y‐axis at A(0, y) and the x‐axis at B(x, 0). Find the smallest area of triangle formed if x and y are positive.

Answer 1

Area = half height times base = xy/2

xy is at a maximum when its derivative = 0

the slope of the line is -y/x

the slope is also 4/(3-x)

set them equal and solve for y

-y/x = 4/(3-x)

-y= 4x/(3-x)

y = -4x/(3-x)

Area = xy/2 = -4x^2/(3-x)/2

take the derivative and set = 0

solve for x

plug that into y=-4x/(3-x)

then multiply xy divide by 2, that's your minimum area of the triangle

to be continued soon

I've run out of time

but I'm pretty confident answer is 24 for minimum area

y=8, x= 6 area =8x6/2 = 48/2=24

If the point had been (3,3) or (4,4) or where the x & y coordinates were the same, the minimum area triangle would be a 45-45-90 degree right triangle, with double the values for x and y intercepts, This problem is close, as just double the (3,4) values to get 3x2=6 for the x intercept and 4x2=8 for the y intercept of the line through (3,4).

Answer 2

The slopes between (3,4) and the intercepts must be the same

(0,y) vs (3,4)

m = (4-y)/(3-0) = (4-y)/3

(x,0) vs (3,4)

m = 4/(3-x)

So

m = (4-y)/3 = 4/(3-x)

(4-y)(3-x) = 12

12 - 4x - 3y + xy = 12

xy - 4x - 3y = 0

y(x-3)-4x = 0

y = 4x/(x-3)

But

area(x,y) = (1/2)xy

area = (1/2)x(4x)/(x-3) <--- substitutes

= (2x^2)/(x-3)

= A(x)

Extrema occurs when derivative is zero:

0 = dA/dx = [(x-3)(4x) - (2x^2)] / (x-3)^2

[(x-3)(4x) - (2x^2)] = 0

4x^2 - 12x - 2x^2 = 0

2x^2 - 12x = 0

x^2 -6x= 0

x(x-6)=0

x=0 or x-6 = 0

x=0 <--- disqualified since x>0

x-6 = 0 --> x = 6

This forces y = 4*6/(6-3) =24/3 = 8

triangle is (0,8), (0,0) (6,0) and has area 24

Answer 3

The slope of the line is -4/(x₀-3)

Use the point-slope formula to get the equation of the line: y= 4(x₀-x)/(x₀-3)

Set x=0 to get y₀ = 4x₀/(x₀-3)

The area of the triangle is .5x₀*4x₀/(x₀-3)

The critical point for x₀ occurs where the derivative (you calculate it & if you can't, let me know) of area with respect to x_o equals 0, i.e.

where x₀=6 and y₀=8