Guest Dem Sum Posted May 29, 2003 Share Posted May 29, 2003 I have 4 of them 1) An open topped cylindrical glass jar is to have a given capacity. Find the ratio of height to diameter if the area of glass is a minimum. 2) The cost per square metre of the sides of an open topped cylindrical tank is twice the cosr per square metre of the bottom. Find the most economical proportions for a tank of given volume. 3) A rectanglar sheet of iron 300 cm wide is to be bent to form a gutter whose cross section is the arc of a circle. What radius will give the gutter the maximum carrying capacity? 4)Find the dimensions of the circular cylinder of largest volume that can be inscribed in a right circular cone of height H and radius of base R. If someone could post their solutions with steps or just steps saying how to do it I'd be very greatful. There were about 10 questions like this but I cannot solve these 4. Thanks. Link to comment Share on other sites More sharing options...
Tom Mattson Posted May 29, 2003 Share Posted May 29, 2003 The best way for you to get help with homework is for you to show how you start, and where you got stuck. I will make some comments that will help you start. Originally posted by Dem Sum 1) An open topped cylindrical glass jar is to have a given capacity. Find the ratio of height to diameter if the area of glass is a minimum. Look up the geometrical formulas for the area and the volume of an open top cylinder. (The volume is the same as that of a closed top cylinder, and the area is the same except you have to subtract off the area of one cap). You are fixing the volume to some constant, and trying to minimize the area. So, at some point you are going to be taking the derivative of the expression for area and setting it equal to zero. From that, you will solve for the ratio of h:d. 2) The cost per square metre of the sides of an open topped cylindrical tank is twice the cosr per square metre of the bottom. Find the most economical proportions for a tank of given volume. This problem is similar to the first one. Set up an expression for the cost of the tank in terms of the diameter and height, and minimize. 3) A rectanglar sheet of iron 300 cm wide is to be bent to form a gutter whose cross section is the arc of a circle. What radius will give the gutter the maximum carrying capacity? You can reduce this problem to that of finding the maximum area under a circular arc for a fixed arc length. Set up the expression for the area, and minimize. 4)Find the dimensions of the circular cylinder of largest volume that can be inscribed in a right circular cone of height H and radius of base R. I would do this with analytic geometry, because you know that the cylinder has to touch the surface of the cone in a circle. You will need to write an expression for the volume of the cylinder using that information. Try to start these, and I will help you through the rough spots. Link to comment Share on other sites More sharing options...
NSX Posted May 30, 2003 Share Posted May 30, 2003 Originally posted by Dem Sum I have 4 of them 1) An open topped cylindrical glass jar is to have a given capacity. Find the ratio of height to diameter if the area of glass is a minimum. 2) The cost per square metre of the sides of an open topped cylindrical tank is twice the cosr per square metre of the bottom. Find the most economical proportions for a tank of given volume. 3) A rectanglar sheet of iron 300 cm wide is to be bent to form a gutter whose cross section is the arc of a circle. What radius will give the gutter the maximum carrying capacity? 4)Find the dimensions of the circular cylinder of largest volume that can be inscribed in a right circular cone of height H and radius of base R. If someone could post their solutions with steps or just steps saying how to do it I'd be very greatful. There were about 10 questions like this but I cannot solve these 4. Thanks. Just like Tom said; but you've got to be careful sometimes. Use the first derivative test to check for critical points, b/c sometimes the derivative may yield a MAX or a MIN; you may also have to check the restrictions placed upon the model. Link to comment Share on other sites More sharing options...
JaKiri Posted May 30, 2003 Share Posted May 30, 2003 Originally posted by NSX Just like Tom said; but you've got to be careful sometimes. Use the first derivative test to check for critical points, b/c sometimes the derivative may yield a MAX or a MIN; you may also have to check the restrictions placed upon the model. f''(x) = + for min, - for max. Link to comment Share on other sites More sharing options...
NSX Posted May 31, 2003 Share Posted May 31, 2003 Originally posted by MrL_JaKiri f''(x) = + for min, - for max. SHould that be f'(x)? f"(x) is the concavity testing.. Link to comment Share on other sites More sharing options...
JaKiri Posted May 31, 2003 Share Posted May 31, 2003 Originally posted by NSX SHould that be f'(x)? f"(x) is the concavity testing.. No. It's not a maximal/minimal turning point if f'(x) doesn't equal zero, and the kind that it is is determined by f''(x) (+ve for min, -ve for max. Oh, and 0 for point of inflection). Link to comment Share on other sites More sharing options...
Guest Brendon Posted June 20, 2004 Share Posted June 20, 2004 Hey, i dont know if any of you, could help me.... i need to find the Derivative, using the DEFINITION OF THE DERIVATIVE: f (x) = 2 ....... x+4 I would be grateful if the complete steps are shown.. tank you Link to comment Share on other sites More sharing options...
Dave Posted June 20, 2004 Share Posted June 20, 2004 Definition of the derivative is: [math]\lim_{h \to 0} \left(\frac{f(x+h)-f(x)}{h}\right)[/math]. So we have [math]\lim_{h\to 0} \frac{\frac{2}{x+4+h} - \frac{2}{x+4}}{h}[/math] as our limit. Notice that if we make the numerator of our fraction into a single fraction, we get the limit as being: [math]\lim_{h\to 0} \frac{2(x+4) - 2(x+4+h)}{(x+4+h)(x+4)h}[/math] Now notice that most of the numerator cancels, giving us: [math]\lim_{h\to 0} \frac{-2}{(x+4+h)(x+4)}[/math] This is a rather trivial limit, yielding the result of [math]\frac{d}{dx} \left(\frac{2}{x+4}\right) = \frac{-2}{(x+4)^2}[/math]. Link to comment Share on other sites More sharing options...
Guest Brendon Posted June 27, 2004 Share Posted June 27, 2004 Thanks alot .. Dave.. i got it.. Link to comment Share on other sites More sharing options...
Dave Posted June 27, 2004 Share Posted June 27, 2004 No problem, the trick with that is just to eliminate the h from the bottom of the limit (if the limit exists ofc). Link to comment Share on other sites More sharing options...
Guest Brendon Posted June 29, 2004 Share Posted June 29, 2004 Hi.. I had to graph this piece wise function which i did, but i had to explain where f is non differentiable but i couldnt do it.... Here's the piece wise function: f(x) = 2x if x < 2 6 - x if x ≥ 2 Link to comment Share on other sites More sharing options...
Dave Posted June 29, 2004 Share Posted June 29, 2004 Obviously the functions 2x and 6-x are differentiable. However, when you get the point x = 2, you have a 'jump' in the function. The easiest way to see that it's not differentiable is to take the limit of both your functions and then show that they're not equal - or just to say that the limit at x=2 does not exist. Link to comment Share on other sites More sharing options...
bloodhound Posted June 29, 2004 Share Posted June 29, 2004 dave, isnt that the condition for continuity? u can have a function which is not differentiable at a point but which is continuous at that point. i.e f(x)=|x| at x=0 Link to comment Share on other sites More sharing options...
Dave Posted June 29, 2004 Share Posted June 29, 2004 If it's discontinuous at a point, then isn't it non-differentiable? Or am I remembering something else? Link to comment Share on other sites More sharing options...
bloodhound Posted June 29, 2004 Share Posted June 29, 2004 oops sorry. i think ur rite. i got my implications the wrong way round. Link to comment Share on other sites More sharing options...
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