[Solved] Differentiation

Apr 2010
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1. The diagram shows an open tank for storing water, ABCDEF. The sides ABFE and CDEF are rectangles. The triangular ends ADE and BCF are isosceles, and Angle AED = BFC = 90deg. The ends ADE and BCF are vertical and EF is horizontal

Given AD = x metres,
a. Show that the area of triangle ADE is 0.25x^2 m^2
b.Given that the capacity of the container is 4000m^3 and that the total are of the two triangular and two rectangular sides of the container is S m^2.

Show that S = (x^2)/2 + {16000(2^1/2)}/x

The question looks straight-forward, i did indeed try it but couldn't figure out the solution.

Plz find attached figure as question 8 in jpeg.

Any help or hint will be much appreciated.

THX
 

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May 2010
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a:

ADE is a right triangle so ue pythagoras' theorum.

\(\displaystyle AD^2 + ED^2 = DE^2\)

AD=ED and DE=x
\(\displaystyle 2AD^2=x^2\)
\(\displaystyle AD=\frac{x}{\sqrt{2}}\)

Area of triangle = 0.5*AD*ED
\(\displaystyle 0.5 * (\frac{x}{\sqrt{2}})^2\)

\(\displaystyle 0.25 * x^2\)


b:
The total area of the container is equal to DC * the area of the triangle. o we just need to find DC.

\(\displaystyle 4000 = DC * 0.25 * x^2\)
But, we know the area of the rectangle satisfies
\(\displaystyle S = 2*DC*DE + 2*0.25 * x^2\)

\(\displaystyle \frac{0.5S -0.25x^2}{DE} = DC\)
\(\displaystyle \frac{0.5S -0.25x^2}{\frac{x}{\sqrt{2}}} = DC\)

Substitute into the other equation and hopefully you can continue from there
 
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Apr 2010
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a:

ADE is a right triangle so ue pythagoras' theorum.

\(\displaystyle AD^2 + ED^2 = DE^2\)

AD=ED and DE=x
\(\displaystyle 2AD^2=x^2\)
\(\displaystyle AD=\frac{x}{\sqrt{2}}\)

Area of triangle = 0.5*AD*ED
\(\displaystyle 0.5 * (\frac{x}{\sqrt{2}})^2\)

\(\displaystyle 0.25 * x^2\)


b:
The total area of the container is equal to DC * the area of the triangle. o we just need to find DC.

\(\displaystyle 4000 = DC * 0.25 * x^2\)
But, we know the area of the rectangle satisfies
\(\displaystyle S = 2*DC*DE + 2*0.25 * x^2\)

\(\displaystyle \frac{0.5S -0.25x^2}{DE} = DC\)
\(\displaystyle \frac{0.5S -0.25x^2}{\frac{x}{\sqrt{2}}} = DC\)

Substitute into the other equation and hopefully you can continue from there
hi, thanks a lot for your response. Could you please explain me why in (b)
volume of container = DC * 0.25* x^2 ????

we dont know the shape of the container in the question, what formula did we apply above?

thanks
 

earboth

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hi, thanks a lot for your response. Could you please explain me why in (b)
volume of container = DC * 0.25* x^2 ????

we dont know the shape of the container in the question, what formula did we apply above?

thanks
1. "Turn" the container until it "stands" on the right triangle. That means you have a right prism with right triangles as base and top areas. As shown by SpringFan25 the base area has a value of \(\displaystyle \frac14 x^2\). (Actually it is a quarter square where the square's side length is x).

2. The volume of a prism is calculated by:

\(\displaystyle V= base\ area \cdot height\)

According to the sketch the height corresponds to the edges DC or AB.

Thus the volume of the prism is calculated by:

\(\displaystyle V = \frac14 x^2 \cdot |DC|\)

3. Since you know the volume of V you are able to calculate the length of DC - as shown by SpringFan25.
 
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Apr 2010
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Thanks a lot SpringFan25 and Earboth

Your help is very much appreciated.

Thanks again.