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geometry of derivative: Stretch of rect. beam varies as product of w, square of d....

Jeane

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Joined
Jun 17, 2017
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22
Here is the question:

"Engineers have determined that the strength s of a rectangular beam varies as the product of the width w and the square of the depth d of the beam; that is, s = kwd^2 for some constant k. Find the dimensions of the strongest rectangular beam that can be cut from a cylindrical log with diameter 48cm."

There is a drawing (which I don't have the technology or know how to present for you); it shows the rectangle inside the circle with the diagonal of the rectangle (which is the diameter of the circle) at 48 cm and the lines/sides to the right angular corner of the rectangle on the circumference, being labelled w(width) and d (depth)

My reasoning so far:

Relevant is Pythagoras' Theorem, making w^2 +d^2 =48^2 and making d^2 the subject d^2=(48^2-w^2), and differentiating the equation for strength ds/dd = 2kwd
Make this equal to zero to find the maximum

I am having difficulty in knowing how to progress with this so w and d can be isolated. I am tempted to substitute d^2 with (48^2 - w^2) into the equation for s, but it doesn't help isolate the terms because there is still the constant k involved.
 
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tkhunny

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Apr 12, 2005
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9,701
Here is the question:

"Engineers have determined that the strength s of a rectangular beam varies as the product of the width w and the square of the depth d of the beam; that is, s = kwd^2 for some constant k. Find the dimensions of the strongest rectangular beam that can be cut from a cylindrical log with diameter 48cm."

There is a drawing (which I don't have the technology or know how to present for you); it shows the rectangle inside the circle with the diagonal of the rectangle (which is the diameter of the circle) at 48 cm and the lines/sides to the right angular corner of the rectangle on the circumference, being labelled w(width) and d (depth)

My reasoning so far:

Relevant is Pythagoras' Theorem, making w^2 +d^2 =48^2 and making d^2 the subject d^2=(48^2-w^2), and differentiating the equation for strength ds/dd = 2kwd
Make this equal to zero to find the maximum

I am having difficulty in knowing how to progress with this so w and d can be isolated. I am tempted to substitute d^2 with (48^2 - w^2) into the equation for s, but it doesn't help isolate the terms because there is still the constant k involved.
Not bad. Just one thing. The point of the Pythagorean relationship is to get you down to one variable.

Given: s(w,d) =k*w*d^2, we have s(w) = k*w*(48^2 - w^2). Find ds/dw from s(w). Multiply it out or use the product rule. Your choice.

'k' is just a constant. Keep it around and don't let it bother you. As far as this problem goes, it probably changes only for different kinds of wood. If you're talking about just one log, you can essentially ignore it.
 
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Jeane

New member
Joined
Jun 17, 2017
Messages
22
k and the last term d

Not bad. Just one thing. The point of the Pythagorean relationship is to get you down to one variable.

Given: s(w,d) =k*w*d^2, we have s(w) = k*w*(48^2 - w^2). Find ds/dw from s(w). Multiply it out or use the product rule. Your choice.

'k' is just a constant. Keep it around and don't let it bother you. As far as this problem goes, it probably changes only for different kinds of wood. If you're talking about just one log, you can essentially ignore it.
I found k cancels out during the calculation but I don't know which equation to plug w into to get d. It equals 16* root 3
 

Jeane

New member
Joined
Jun 17, 2017
Messages
22
got it

I found k cancels out during the calculation but I don't know which equation to plug w into to get d. It equals 16* root 3
It's the d^2 equation So I'm Ok now Thanks for your help
 
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