supplement #9, #11
For #9 i set it up as y=(x-315)^2+630. I thought to look for (115,y) on the graph since this is where it would have a width of 400 and to see if y was greater than or equal to 400 because then the cube would fit through. However, I got a concave up parabola and was confused. Any different ideas on how to approach this? Also, for number 11 on the supplement I am confused on how to set it up. Thanks
posted by Carrie Desmond at 3:28 PM
2 Comments:
I'm not sure I did 9 completely right, but for my equation I got f(x)= (-2/315)*(x-315)^2+630. Don't forget that you have to find a value for "a" if f(x)=a*(x-h)^2+k. You plugged in the coordinates of the vertex, but you then have to plug the coordinates of a known point in for x and y to solve for a. Since you know (0,0) is on the graph (this is only the case of the vertex is (315,630), but you set the problem up the same way I did so it works), you can plug these coordinates in and you get an "a" value of -2/315. This makes the parabola concave down.
11 confused me too. Do we have to taken the volume of the individiual sides into account??! I don't have my solution with me since Mr. Karafiol collected it, but I think ignored the part that said "the glass for the sides will be two times as thick as the glass for the bottom" and solved it without that info. Haha, obviously I can't help.
For #11, i think "the glass for the sides will be two times as thick as the glass for the bottom" means you should time the surface area of the sides by two, then you can stick the two surface of the sides together to get two times as thick as the bottom. Does it make sense? Since volume=x^2*h, h=90000/x^2. The i did y=x^2+2*4*(x*90000/x^2), graph the funcion and find the minimun point.
I hope it can help you.
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