# geometry of derivative: (a) Draw region enclosed by parab. y^2=4ax & latus rectum x=a

#### Jeane

##### New member
geometry of derivative: (a) Draw region enclosed by parab. y^2=4ax & latus rectum x=a

Having difficulty with the formulation of the equation from which the derivative is to be gained again.
Here is the question: "(a) Draw a diagram showing the region enclosed between the parabola y^2=4ax and its latus rectum x=a. (b) Find the dimensions of the rectangle of maximum area that can be inscribed in this region."

I can't draw here the graph but know it's a parabola on its side opening out on the RHS with the vertex at the origin being cut by the latus rectum from x=a
the latus rectum from point of contact to point of contact with the parabola is 4a long.

Pertinent is the area of the rectangle is Base x Height. But discerning these algebraically is difficult for me. One side of the rectangle lies along the latus rectum
and thus would be in terms of a and the other side would involve terms that come from the point of contact with the parabola but that is as far as I've got with my thinking. I have wondered if looking at it the other way round might be better. Thanks.

#### stapel

##### Super Moderator
Staff member
Having difficulty with the formulation of the equation from which the derivative is to be gained again.
Here is the question: "(a) Draw a diagram showing the region enclosed between the parabola y^2=4ax and its latus rectum x=a. (b) Find the dimensions of the rectangle of maximum area that can be inscribed in this region."

I can't draw here the graph but know it's a parabola on its side opening out on the RHS with the vertex at the origin being cut by the latus rectum from x=a
the latus rectum from point of contact to point of contact with the parabola is 4a long.

Pertinent is the area of the rectangle is Base x Height. But discerning these algebraically is difficult for me. One side of the rectangle lies along the latus rectum
and thus would be in terms of a and the other side would involve terms that come from the point of contact with the parabola but that is as far as I've got with my thinking. I have wondered if looking at it the other way round might be better. Thanks.
Draw the enclosed region. Draw a rectangle inside this region, with the left-hand corners touching the parabola. (Don't bother with "finding" the maximal rectangle. You're just using the rectangle to show you the logic.)

i. The area of the rectangle is twice the area of the above-the-x-axis portion, being the upper half of the rectangle. How does this simplify your work?

ii. The area of the upper half is the value of the base (that is, the enclosed width), multiplied by the value of the height. Looking at the picture, what expression must stand for the value of the height of the upper-half?

iii. You know the x-value of the right-hand end of the upper-half. What expression or variable stands for the value of the left-hand end?

iv. What expression then must you maximize?

#### Jeane

##### New member

Draw the enclosed region. Draw a rectangle inside this region, with the left-hand corners touching the parabola. (Don't bother with "finding" the maximal rectangle. You're just using the rectangle to show you the logic.)

i. The area of the rectangle is twice the area of the above-the-x-axis portion, being the upper half of the rectangle. How does this simplify your work?

ii. The area of the upper half is the value of the base (that is, the enclosed width), multiplied by the value of the height. Looking at the picture, what expression must stand for the value of the height of the upper-half?

iii. You know the x-value of the right-hand end of the upper-half. What expression or variable stands for the value of the left-hand end?

iv. What expression then must you maximize?

i. No dealing with negative values that complicate calculations
ii.y=square root 4ax
iii. Bringing in another variable it would be a-b where x=b is the LHS of the rectangle
iv.Must maximise Area=(a-b)*(square root 4ax)
v. Now I'm stuck

#### Jeane

##### New member
My term b

i. No dealing with negative values that complicate calculations
ii.y=square root 4ax
iii. Bringing in another variable it would be a-b where x=b is the LHS of the rectangle
iv.Must maximise Area=(a-b)*(square root 4ax)
v. Now I'm stuck
Is my term b really x=(y^2)/4a?

#### stapel

##### Super Moderator
Staff member
i. No dealing with negative values that complicate calculations
More to the point, you can deal only with the top half, and then multiply the half-rectangle result by 2 to get the full-rectangle result.

ii.y=square root 4ax
For some value of x (to be determined), yes.

iii. Bringing in another variable it would be a-b where x=b is the LHS of the rectangle
Using "a" (from the parabola formula) as the location (along the x-axis) of the focus, yes, this will give you the value of the left-hand side's x-value. However, this assumes that you know what that value is, that "b" is fixed. Since "b" depends upon "x", surely "x" should be used?

iv.Must maximise Area=(a-b)*(square root 4ax)
Replace "b" with "x", and I'd agree.

v. Now I'm stuck
Um... "derivatives", maybe...? And solve for "x" in terms of "a"...?

#### Jeane

##### New member

i. No dealing with negative values that complicate calculations
More to the point, you can deal only with the top half, and then multiply the half-rectangle result by 2 to get the full-rectangle result.

ii.y=square root 4ax
For some value of x (to be determined), yes.

iii. Bringing in another variable it would be a-b where x=b is the LHS of the rectangle
Using "a" (from the parabola formula) as the location (along the x-axis) of the focus, yes, this will give you the value of the left-hand side's x-value. However, this assumes that you know what that value is, that "b" is fixed. Since "b" depends upon "x", surely "x" should be used?

iv.Must maximise Area=(a-b)*(square root 4ax)
Replace "b" with "x", and I'd agree.

v. Now I'm stuck
Um... "derivatives", maybe...? And solve for "x" in terms of "a"...?
the derivative is too complex to solve. I used the product rule

. . .Area=(a-x) (square root 4ax)

. . .dA/dx= u*dv/dx +v*du/dx

where u= (a-x) and du/dx=-1, v=square root 4ax and dv/dx= 1/(square root 4ax)*2

so the derivative becomes

. . .(a-x)/2*(square root 4ax)- (square root 4ax)

. . . = (a-x)/2*(square root 4ax) -(square root 4ax) *(2*(square root 4ax)/2*(square root 4ax))

. . .= (a-x-8ax)/(square root 4ax)*2

and it's here that I'm stopped. I could have made a mistake This looks rather clumsy. thanks for your reply if you still can.

Last edited by a moderator:

#### stapel

##### Super Moderator
Staff member
the derivative is too complex to solve. I used the product rule

. . .Area=(a-x) (square root 4ax)

. . .dA/dx= u*dv/dx +v*du/dx

where u= (a-x) and du/dx=-1, v=square root 4ax and dv/dx= 1/(square root 4ax)*2

so the derivative becomes

. . .(a-x)/2*(square root 4ax)- (square root 4ax)

. . . = (a-x)/2*(square root 4ax) -(square root 4ax) *(2*(square root 4ax)/2*(square root 4ax))

. . .= (a-x-8ax)/(square root 4ax)*2

and it's here that I'm stopped. I could have made a mistake This looks rather clumsy. thanks for your reply if you still can.
Your formatting is ambiguous. I've broken the stream-of-consciousness math into separate lines, but I'm still confused.

The derivative, in the first pass, should have been as follows:

. . . . .$$\displaystyle A(x)\, =\, (a\, -\, x)\, 2\sqrt{\strut ax\,}$$

. . . . .$$\displaystyle A'(x)\, =\, (-1)\,(2)\, \sqrt{\strut ax\,}\, +\, (a\, -\, x)\, (2)\, \dfrac{a}{2\, \sqrt{ax\,}}$$

. . . . . . . . .$$\displaystyle =\, -2\, \sqrt{\strut ax\,}\, +\, \dfrac{2a\,(a\, -\, x)}{2\, \sqrt{ax\,}}$$

Set this equal to zero. Solve the resulting linear equation for x in terms of "a".

#### Jeane

##### New member
Thanks that's enough Cheers!

Your formatting is ambiguous. I've broken the stream-of-consciousness math into separate lines, but I'm still confused.

The derivative, in the first pass, should have been as follows:

. . . . .$$\displaystyle A(x)\, =\, (a\, -\, x)\, 2\sqrt{\strut ax\,}$$

. . . . .$$\displaystyle A'(x)\, =\, (-1)\,(2)\, \sqrt{\strut ax\,}\, +\, (a\, -\, x)\, (2)\, \dfrac{a}{2\, \sqrt{ax\,}}$$

. . . . . . . . .$$\displaystyle =\, -2\, \sqrt{\strut ax\,}\, +\, \dfrac{2a\,(a\, -\, x)}{2\, \sqrt{ax\,}}$$

Set this equal to zero. Solve the resulting linear equation for x in terms of "a".
Thanks that's enough Cheers!