Max and Min (1 Viewer)

[fullonoob]

Find the value of a for which the area of triangle formed by the tangent and the coordinate axes will be minimum.
Equation of tanget at P : y = a^2 - 2ax + 4
Tangent : P (a , 4-a^2)

Thanks

 

[Trebla]

Find the value of a for which the area of triangle formed by the tangent and the coordinate axes will be minimum.
Equation of tanget at P : y = a^2 - 2ax + 4
Tangent : P (a , 4-a^2)

Thanks

Equation of line (assuming it is correct):
y = a²-2ax+4
y-intercept occurs at y = 4
x-intercept occurs at x = (a²+4)/2a
Hence area of triangle is:
A = 1/2 (4) (a²+4)/2a
= (a²+4)/a
= a + 4/a
Want to minimise A (> 0) with respect to a
dA/da = 1 - 4/a²
dA/da = 0 => a = 2 or - 2
d²A/da² = 8/a³
The second derivative is positive (for a minimum) when a > 0, hence choose a = 2 which gives a local minimum. Since A > 0, then domain is: a > 0 thus there are no other turning points in this domain, so the local minimum is also the absolute minimum.

 

[fullonoob]

thats what i did
but the answer is 2/ root 3 which is 2/3 x root 3

 

[fullonoob]

Another question i cant do..should be easy
The point P (x , y) lies on the parabola y = x ^2
Show that the distance from P to the line x-y-1 = 0 is
1/2 (times) root 2 (x^2 - x + 1)

Sorry I'm noob

 

[scardizzle]

so (x,y) lie on the parabola y = x^2

therefore (x,y) can be written as (x,x^2)

assuming the q is talking about its perpendicular distance to that line

distance = | x - x^2 -1| x (1 + 1)^-1/2

= |x^2 -x +1| x 2^-1/2 (dividing by -1 this wont affect the answer since there are absolute value brackets)

= 1/2 (times) root 2 (x^2 - x + 1) (rationalizing the denominator)