Archive
20th of March 2017
Mathematics
Q:

A metal box with a square base and vertical sides is to contain 1024c m cubed. The material for the top and bottom costs Rs 5 perc m squaredand the material for the sides costs Rs. 2.50 perc m squared. Find the least cost of the box.

Ahuja8087 Ahuja - CBSE - Class XII Science

Monday, March 20, 2017 at 17:21:PM

A:

begin mathsize 16px style Let space the space side space of space the space square space be space straight x space and space the space vertical space side space be space straight y.
So comma space volume space of space the space box equals straight x squared straight y
rightwards double arrow 1024 equals straight x squared straight y
rightwards double arrow straight y equals 1024 over straight x squared space.... left parenthesis straight i right parenthesis
Area space of space the space sides equals 4 xy space and space area space of space the space top space and space bottom space equals space 2 straight x squared
Let space the space total space cost space be space straight C.
Given space that space the space material space for space the space top space and space bottom space costs space Rs. space 5 space per space cm squared space and space that space for space the space sides space costs space Rs. space 2.50 space space per space cm squared
So comma space straight C equals 5 left parenthesis 2 straight x squared right parenthesis plus 2.5 left parenthesis 4 xy right parenthesis
rightwards double arrow straight C equals 10 straight x squared plus 10 xy
rightwards double arrow straight C equals 10 straight x squared plus 10 straight x open parentheses 1024 over straight x squared close parentheses space... left parenthesis from space left parenthesis straight i right parenthesis right parenthesis
rightwards double arrow straight C equals 10 straight x squared plus 10240 over straight x squared
Now space differentiationg space wrt space straight x comma space we space get

straight C apostrophe equals 20 straight x minus 10240 over straight x equals fraction numerator 20 straight x squared minus 10240 over denominator straight x end fraction
For space the space critical space point comma space straight C apostrophe space equals space 0
rightwards double arrow fraction numerator 20 straight x squared minus 10240 over denominator straight x end fraction equals 0
Using space this comma space solve space for space straight x.
differentiate space straight C apostrophe space again comma space and space substitute space the space value space of space straight x
You space will space get space straight C apostrophe apostrophe greater than 0
So comma space the space cost space would space be space minimum.
Substitute space the space value space of space straight x comma space in space straight C space to space find space the space least space cost. end style

Monday, March 20, 2017 at 17:45:PM