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Calculus and a Trapezium (maximum)

Hi guys, ive been staring at this question for a while now and im hitting a brick wall every time. Some help would be fantastic.


I have this trapezium, the question is, "The cross section of an open channel is a trapezium, with base 6cm and slopping sides each 10cm long. calculate the with of the open top so that the cross sectional area shall be a maximum"

Here is my working so far up until my brick wall.

x2+h2=102[br]h2=102−x2[br]h=102−x2x^2+h^2=10^2[br]h^2=10^2-x^2[br]h={\sqrt{10^2-x^2}}

The area of a triangle = hx/2
The area of a rectangle = 6h

Total area = 2hx2+6h2{\dfrac{hx}{2}}+6h
= hx+6hhx+6h
= (x+6)h(x+6)h




using h=102−x2h={\sqrt{10^2-x^2}} in total area calculation

total area = (x+6)∗102−x2(x+6)*{\sqrt{10^2-x^2}}

And this is were i hit a brick wall, i know that i need to differentiate this (x+6)∗102−x2(x+6)*{\sqrt{10^2-x^2}} and then set the differential to zero to get an answer that i can use to calculate xx, but not sure how to do it.

Could anyone help me with this?

Cheers Ramjam
Original post by Ramjam
Hi guys, ive been staring at this question for a while now and im hitting a brick wall every time. Some help would be fantastic.


I have this trapezium, the question is, "The cross section of an open channel is a trapezium, with base 6cm and slopping sides each 10cm long. calculate the with of the open top so that the cross sectional area shall be a maximum"

Here is my working so far up until my brick wall.

x2+h2=102[br]h2=102−x2[br]h=102−x2x^2+h^2=10^2[br]h^2=10^2-x^2[br]h={\sqrt{10^2-x^2}}

The area of a triangle = hx/2
The area of a rectangle = 6h

Total area = 2hx2+6h2{\dfrac{hx}{2}}+6h
= hx+6hhx+6h
= (x+6)h(x+6)h




using h=102−x2h={\sqrt{10^2-x^2}} in total area calculation

total area = (x+6)∗102−x2(x+6)*{\sqrt{10^2-x^2}}

And this is were i hit a brick wall, i know that i need to differentiate this (x+6)∗102−x2(x+6)*{\sqrt{10^2-x^2}} and then set the differential to zero to get an answer that i can use to calculate xx, but not sure how to do it.

Could anyone help me with this?

Cheers Ramjam


Differentiate using the product rule.
Reply 2
Original post by brianeverit
Differentiate using the product rule.


Okay ive got −x(x+6)100−x2+100−x2-{\dfrac{x(x+6)}{\sqrt{100-x^2}}}+{\sqrt{100-x^2}}
Original post by Ramjam
Okay ive got −x(x+6)100−x2+100−x2-{\dfrac{x(x+6)}{\sqrt{100-x^2}}}+{\sqrt{100-x^2}}


Add the fractions and set the numerator =0
Reply 4
Original post by TenOfThem
Add the fractions and set the numerator =0

can you show me what you mean please?
Original post by Ramjam
can you show me what you mean please?


Put it all over a common denominator of 100−x2\sqrt{100-x^2}
so you will have −x(x+6)+(100−x2)100−x2\frac{-x(x+6)+(100-x^2)}{\sqrt{100-x^2}}
Now put the numerator =0 and a solve for x
(edited 9 years ago)
Reply 6
Original post by brianeverit
Put it all over a common denominator of 100−x2\sqrt{100-x^2}
so you will have −x(x+6)+(100−x2)100−x2\frac{-x(x+6)+(100-x^2)}{\sqrt{100-x^2}}
Now put the numerator =0 and a solve for x

I simplified the numerator to make it easier which i got as -2x^2-6x+100,
am i right in thinking this has to be solved using the quadratic formula?

meaning that (a=2, b=6, c=-100)
which loads to be getting

resulting in = 5.73 and -8.73

and the only valid answer there being x=5.73

is this correct??
(edited 9 years ago)
Original post by Ramjam
I simplified the numerator to make it easier which i got as -2x^2-6x+100,
am i right in thinking this has to be solved using the quadratic formula?

meaning that (a=2, b=6, c=-100)
which loads to be getting

resulting in = 5.73 and -8.73

and the only valid answer there being x=5.73

is this correct??


It looks o.k. to me.
However, don''t foget that this is not the required answer.
(edited 9 years ago)
Reply 8
yeh my final answer will be 6+5.73+5.73

thanks for the help

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