anonymous
  • anonymous
Let's construct a box whose base length is 3 times its base width and will have no lid. We only have 64 m2 of material to use. Determine the dimensions of the box that will give the largest volume. ___width? ___height? ___length? What is the volume of the box? ANY HELP ON HOW TO START THIS PLEASE??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
we know that the surface area has to equal 64m^2 right?
amistre64
  • amistre64
our dimensions can be l,w,h were l = 3w
amistre64
  • amistre64
surface area = 3w(w) + 2(hw) + 2(3wh) = 64

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amistre64
  • amistre64
volume of the box = 3w(w)h
amistre64
  • amistre64
64 = 3w^2 +2wh +6wh 64 = 3w^2 +8wh solve for "h" 64 - 3w^2 = 8wh 64-3w^2 -------- = h 8w
amistre64
  • amistre64
use this "value" of h in the formula for the volume: V = 3w^2 [(64-3w^2)/8w]
amistre64
  • amistre64
V = (192w^2 - 9w^4)/8w ; now simplify
amistre64
  • amistre64
V = (192x -9w^3)/8 find the derivative of this Volume function to determine critical numbers
amistre64
  • amistre64
8(192-27w^2) ------------- 8(8) 192-27w^2 ---------- solve for zero 8 192 - 27w^2 = 0 gonna be tricky for me without pencil and paper :)
amistre64
  • amistre64
got it.. 3(64-9w^2) = 0 64-9w^2 = 0 (8+3w)(8-3w) = 0 w = -3/8 or w = 3/8 since a negative width is meaningless; lets use w = 3/8
amistre64
  • amistre64
recall that l = 3w: l = 3(3/8) = 9/8
amistre64
  • amistre64
64-3w^2 -------- = h 8w 64 - 3(3/8)^2 ------------ = h 8(3/8) 64 - (27/64) ------------ = h 3
amistre64
  • amistre64
4069/64 -------- = h 3 4069/192 = h might need to reduce that....

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