Posted in March 18, 2009 ¬ 3:13 pmh.adminNo Comments »
Open Top , Cm2 to a maximum of 1300 open squares and basic data available Through to find the box?
Cm2 of the 1300 myeondoeneun Base As a base to create an open top box with a square and the box Find the largest possible volume. The volume = cm3. For help Thank you for you really!
Let both sides of the rectangle Wed of the length of the base. Let H is the height of the box. Areas ^ 2 (bass), and * The H-(for both sides said 4 *). We are familiar with the area It is because of the perspective of 1300 s and 1300 s ^ 2 + h we now in the 4sh I'm going to put = 1300 ................... Subtract both the s ^ reuljigo Two equations,: 4sh = 1300 - 1300 ^ 2's ............. Now (= (divide both sides of I h, 2 H Order by separating affection - s ^ 2) / (4 * these)) "- ------------------ first major Point is, we know that the volume of H * response * (if But the problem with this email me at, s only your e-mail Response, something that all the top post since almost every Asker How do I clear e-mail if you'll accept.) V I = a [s ^ 2 * h = v points to the important and the beginning h. Instead, we can * s ^ 2 and ((1300 - s ^ 2) / Jung )].... .... s righteousness 21 divided by the denominator molecular reulha S ^ s get out of. v = (1 / 4) * (response * (1300 - s ^ 2) = (1 / 4) * (1300s - 3 s ^ a) Now we differentiate the volume of the DV response Terms / ds a = (1 / 4) * (1300 - 3S ^ 2 ).............. as s 0 for the solution as well as We have a problem, even all these settings. (1 / 4) * (1300 - 3S as ^ 2) = ^ 2 = 1300 s in the 3S 0 ................ Multiply both sides 4 1300 - 3S ^ 2 = 0 1300 = 3S to the ^ 2 ^ 2 = (1300 / 3) .. ................... We now find s righteousness, and take the square root of both sides. <------ The second important point.'s = Sqrt (1300 / 3 )................ In Algebra It will be a plus or minus working with While we are here so it's a negative side to Can not be. Now go back and derive the second derivative.'s DV / DS for = (1 / 4) * (1300 - 3S Ro ^ 2) d of two 2v/ds ^ ^ 2 = (1 / 4) * (0 - 6S). This is negative, So we have to know that we do best. Since H V * ='s ^ 2, the point is that it is important to us (1300 / s or 3) ^ 22 (Reference.) H. Then we have Find me on the first major Copy point. h, = ((1300 - s ^ 2) / (4 * These)) <------------------- First Important point, so v = 2 = 2 ^ 2 ^ 2 * s neunyiya of H * ((1300 - s ^ 2) / S in the denominator of the divide positive )............ molecules s ^ 2 Therefore, the molecular Now we can go: v and = ((1300s -) ^ 3 s of 4, so instead )............... We know that s = Sqrt (1300 / 3) that the other equation: v and = ((1300 *'s Sqrt (1300 / 3)) - (1300 / 3) ^ (3 / 2)) / 4 ..................... Now we have (simplified out by Sqrt consider the (1300 / 3) v a = ((Sqrt (1300 / 3) / 4) * 1300 - (1300 / 3 ))..... Finally, to simplify 3 ...........( note that (2 / 3) * (1 / 4) = (1 / 6) v and = ((Sqrt (1300 / 3) / 4) * ((2 / 3) * 1300) = [((Sqrt (1300 /) A / 6) * 1300), then] the largest possible volume of cm [((Sqrt (1300 / 3) / 6) * 1300)] cm ^ 3 in simplified form <------------ Marked 4510.276 This is about 4700 cm ^ 3 ^ 3 The estimation problem before my neighbors. So the answer is That almost certainly ..
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