Finding dimensions of a rectangular box to optimize volumeA rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of 16 f t 2 . Compute the dimensions of the box that will maximize its volume.I am going wrong somewhere but I can't see where. I have set 2 ( 3 h + 3 w + h w ) = 16 and my optimization equation is v = h w l. I set w = 8 − 3 h 3 + h and plug into the optimization equation. Calculating v' gives me 9 h 2 − 84 h + 72 ( 3 + h ) 2 ,but setting that equal to zero does not give me the answer. I need for h, which is 1.

Question

Finding dimensions of a rectangular box to optimize volumeA rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of   16  f      t    2  . Compute the dimensions of the box that will maximize its volume.I am going wrong somewhere but I can&#039;t see where. I have set   2  (  3  h  +  3  w  +  h  w  )  =  16 and my optimization equation is   v  =  h  w  l. I set   w  =            8      −      3      h              3      +      h       and plug into the optimization equation. Calculating v&#039; gives me             9              h        2            −      84      h      +      72                      (        3        +        h        )            2      ,but setting that equal to zero does not give me the answer. I need for h, which is 1.

Abbigail Vaughn · Accepted Answer

Explanation:The equation is going wrong you forgot to subtract the top so itll be   2  (  3  b  +  b  h  +  3  h  )  −  3  b  =  16 here b=breadth and h=height . Now you proceed by the way as you did and you will get an equation in h for volume diffrentiate it and put it to be 0. Thus we get eqn             48      h      −      18              h        2                    3      +      2      h       so on differentiating wrt h we get a quadratic as       h    2    +  2  h  −  4  =  0 thus   h  =            −      1      +      ,      −              4        +        16              2   so we take positive sign for max area thus its   −  1  +      5   now you can get breadth and you are done with the sum.

comAttitRize8 · Accepted Answer

Step 1Did you forget that the top of the box is open? The surface area is   2  (  3  h  +  h  w  )  +  3  w  =  16  , so that   w  =            16      −      6      h              3      +      2      h        ..Step 2The volume of the box is   V  =  3  h  w  =  3  h            16      −      6      h              3      +      2      h        .If you differentiate this with respect to h, and set it equal to 0, you get the h for which the volume is maximized.

A rectangular box (base is not square) with an open top must have a length of 3ft, and a surface area of 16ft^2. Compute the dimensions of the box that will maximize its volume.

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