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An open box with asquare base is to be made out of a given quantity of card board of area c2square units. Show that the maximum volume of the box is `(c^3)/(6sqrt(3))`cubic units. |
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Answer» As base of the box is square, So, area of the box `= a^2+4ah` Here, `a` is the side of square base and `h` is the height. Now, it is given that, `c^2 = a^2+4ah` `=>h =1/4 [(c^2-a^2)/a]` Now, Volume of the box`(V) = a*a*h = a^2h` `=>V = 1/4 [(c^2-a^2)/a]a^2` `=>V = 1/4[ac^2-a^3]->(1)` `=>(dV)/(da) = 1/4[c^2 - 3a^2]` Now, for maximum volume, `(dV)/(da) = 0` `=>1/4[c^2 - 3a^2] = 0` `=>c^2 = 3a^2 => a = c/sqrt3` Now, Putting value of `a` in (1), `=>V_max = 1/4[(c/sqrt3)(c^2) - (c/sqrt3)^3]` `=>V_max = c^3/(6sqrt3)` So, maximum volume of the box is `c^3/(6sqrt3)` cubic units. |
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