Rapper 0 Posted June 1, 2016 (edited) A cylinder with radius r=0.75 m and height h = 1 m has to be covered by a cone. Find the minimal area for the cone. Surface Area of Cone: pi*r^2 + pi*r*L So, I figured out: H= Height of the cone R= Radius of the cone (H-1)/0.75 = H/R H=-4R/(-4R+3) To calculate L, I used the Pythagoras & substituted H from above: L = sqrt(R^2+H^2) L = sqrt(R^2+(-4R/(-4R+3)) L = (sqrt(16R^4-24R^3+25R^2)/(4R-3) I see I'm heading towards something really complicated... Could you please help me find the right way to do it? I have just started doing this. Edited June 1, 2016 by Rapper 0 Share this post Link to post Share on other sites

Daedalus 329 Posted June 2, 2016 (edited) Could you please help me find the right way to do it? I have just started doing this. I took the time and solved this problem. However, it does sound a lot like homework. So, I'm going to guide you through how to set up the problem, but you will need to do the math. Without considering the area for the base of the cone, what can you determine about the remaining area? If we look at a 2d cross section of the cone and cylinder cut through the center, we get the following image: Now, the equation for the area of the cone minus the base, or the lateral surface area of the cone, is equal to [math]\pi[/math] times the radius of the cone [math]\left(r+x\right)[/math] times the length of the slant or the hypotenuse of the large triangle shown in the image [math]\left(\sqrt{x^2+h^2}+\sqrt{y^2+r^2}\right)[/math]. [math]A_L(x,y)=\pi r l[/math] Plugging in all of the variables yields the following equation: [math]A_L(x,y)=\pi \left(r+x\right) \left(\sqrt{x^2+h^2}+\sqrt{y^2+r^2}\right)[/math] Now, there is a trigonometric relationship that you need to use. From the image, we know that: [math]\frac{r}{y}=\frac{x}{h}[/math] Using this trigonometric relationship, you should be able to solve for [math]y[/math] in terms of [math]x[/math] (or vice-versa) and substitute that result into the equation for the lateral surface area of the cone and arrive at an equation in one variable ([math]r[/math] and [math]h[/math] are constants). From there, you need to find the minima of this equation by taking the derivative and setting it equal to zero and solving for [math]x[/math] or [math]y[/math] (whichever variable you used during substitution). Once you know the value of the singled out variable ([math]x[/math] or [math]y[/math]) , you can solve for the other variable and arrive at the answer. I've set you on the path, now all you need to do is the work. I'll help if you get stuck, and I can even check your results, but it would be best if your worked the math. Edited June 2, 2016 by Daedalus 0 Share this post Link to post Share on other sites

imatfaal 2480 Posted June 2, 2016 ! Moderator Note I agree with Daedalus - this looks a bit like homework. You have been given an excellent start - now it is your turn to use the ideas given to develop an answer. Members will be happy to give advice, corrections, and feedback - but we don't do answers 0 Share this post Link to post Share on other sites

Robittybob1 17 Posted June 3, 2016 (edited) A cylinder with radius r=0.75 m and height h = 1 m has to be covered by a cone. Find the minimal area for the cone. Surface Area of Cone: pi*r^2 + pi*r*L .... A search of the web [latex]A=\pi {r}({r}+\sqrt{{h_c}^2+r^2})[/latex] Let r = radius of base [latex]h_c[/latex] = height of cone Both equations seem to be equivalent. It seems logical that since L > r then the area of the side of the cone is always greater than the area of the base. Since L > r then area of side will be minimum when L is minimum. Edited June 3, 2016 by Robittybob1 0 Share this post Link to post Share on other sites