geometry of the derivative (1 Viewer)

[jellybelly59]

1.Two sides of a rectangle lies along the x and y axes. The vertex opposite the origin is in the first quadrant and les on 3x + 2y = 6 . What is maximum area of the rectangle?

2.The point P(x,y) lies on the parabola y=x^2
(a) Show that the distance from P to the line x-y-1 =0 is [root2.(x^2-2x+1)] /2
(b) Hence find P for which the distance is minimum.

 

[x.Exhaust.x]

1.Two sides of a rectangle lies along the x and y axes. The vertex opposite the origin is in the first quadrant and les on 3x + 2y = 6 . What is maximum area of the rectangle?

2.The point P(x,y) lies on the parabola y=x^2
(a) Show that the distance from P to the line x-y-1 =0 is [root2.(x^2-2x+1)] /2
(b) Hence find P for which the distance is minimum.

1.

Rearrange y to:

y= (-3/2)x+3

Let the area of the rectangle be: xy

We know y, so we substitute it into the area of the rectangle.

Therefore:

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

A'=-3x+3
= -3(x-1)

x=1 (sub this co-ordinate back into the orginal to get y)
y=3/2

Therefore maximum area of rectangle is 3/2

 

[Timothy.Siu]

1.Two sides of a rectangle lies along the x and y axes. The vertex opposite the origin is in the first quadrant and les on 3x + 2y = 6 . What is maximum area of the rectangle?

2.The point P(x,y) lies on the parabola y=x^2
(a) Show that the distance from P to the line x-y-1 =0 is [root2.(x^2-2x+1)] /2
(b) Hence find P for which the distance is minimum.

first one, i labeled the base of the rectangle as P, then the height would be 3-1.5p
(this is because the top vertex lies on the line)

Area of rectangle is=3p-1.5p^2

A'=3-3p
=0 when p=1
sub in p=1 into area of rectangle
A=1.5

2nd one,
distance from P to line implies perpendicular distance.
so

D=|x-y-1|/root2
=|x-x^2-1|/root 2 (since y=x^2)

=x^2-x+1/root2=root2(x^2-x+1)/2 (since -x^2+x-1 is always less than 0) (lol my answer is different to the answers )

D'=root2(2x-1)/2
=0 when x=1/2

 

[jellybelly59]

timothy did u manage to get point p?

 

[Timothy.Siu]

oh, i wasn't reading the question

i got (1,1) but it might be wrong because my other part was wrong

edit: actually i got (0.5, 0.25) but if i used the results from the 2nd part it would be (1,1)

 

[jellybelly59]

oh, i wasn't reading the question

i got (1,1) but it might be wrong because my other part was wrong

edit: actually i got (0.5, 0.25)

yerh (.5 , .25 ) is right

 

[Timothy.Siu]

yerh (.5 , .25 ) is right

ah cool

 

[jellybelly59]

got a new question i need help with:

The point A lies on the positive half of the x-axis, the point B lies on the positive half of the y axis and the interval AB passes through the point P (5,3). Find the coordinates of A and B so that triangle AOB has minimum area.

 

[Timothy.Siu]

got a new question i need help with:

The point A lies on the positive half of the x-axis, the point B lies on the positive half of the y axis and the interval AB passes through the point P (5,3). Find the coordinates of A and B so that triangle AOB has minimum area.

this is similar to one of the questions in my 2unit trials

A(a,0) B(0,b)

equation of line AB:y-b=-b/a(x-0)

sub in point (5,3) and rearranging u get a=5b/b-3

area of triangle is 1/2ab
sub in a A=0.5x(5b^2/b-3)
A'=0.5(10b(b-3)-5b^2)/(b-3)²

A'=0 when 10b^2-30b-5b^2=0
5b(b-6)=0 b=6 (ur meant to check that this is minimum)
then a=10

btw, where did u get this question from?

 

[jellybelly59]

cambridge

 

[Timothy.Siu]

cambridge

ok cool, looks like our school basically copied it.