calculus question dimensions

Question

a box with an open top has a square base and four sides of equal height. The volume of the box is 441 feet cubed. the height is 2 feet greater than both the length and the wdth. If the surface area is 301 feet squared, what are the dimensions of the box?

Answer 1

open top was ignored an oversight in following calculations.

it doesn't change the volume but does change the surface area

SA = 2hw +2hL + wL=301, and if w=L=x V=hx^2=441, h=441/x^2, then

301 = 4hx +x^2

x^2 +4(441/x^2)x =301

x^2 + 1764x -301 = 0 (x-0.17)(x+1764.17) = 0

x = about .17 = width = Length but then

h = 2.17 and V= 2.17(.17)^2 <<411

There's an inconsistency in the given information, no real solution exists. DNE, Null set

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unless you like imaginary or negative dimensions for boxes

V = hwL = 441 ft^3, SA = 2(hw + hL + wL) = 301 ft^2, h = 2+L+w

h = 441/wL = 2+L+w

441 = 2wL + wL^2 + Lw^2

wL^2 + (2w +w^2)L -441 =0

L = (-2w -w^2)/2w +/-(1/2w)sqr((2w+w^2)^2 + 4w(441))

L = -1 - w/2 +/- (1/2w)sqr(w^4 +4w^3 +4w^2 + 1764)

substitute that into the L below

hw + hL +wL = 100.5

(2+L+w)w + (2+L+w)L = 100.5

(2+w -1-w/2 +/1/2w)sqr(w^4 +4w^3 +4w^2 +1764) +(2+w -1-w/2+/-(1/2)sqr(w^4 +4w^3 +4w^2 +1764)(-1-w/2 +/- (1/2w)sqr(w^4 +4w^3 +4w^2 + 1764)

now you reduced it to one equation one unknown, solve for w= width, graph it find the x intercepts if any, x intercepts = width, but look out for possible extraneous solutions

Answer 2

V = hwL = 441 ft^3, SA = 2(hw + hL + wL) = 301 ft^2, h = 2+L+w

h = 441/wL = 2+L+w

441 = 2wL + wL^2 + Lw^2

wL^2 + (2w +w^2)L -441 =0

L = (-2w -w^2)/2w +/-(1/2w)sqr((2w+w^2)^2 + 4w(441))

L = -1 - w/2 +/- (1/2w)sqr(w^4 +4w^3 +4w^2 + 1764)

substitute that into the L below

hw + hL +wL = 100.5

(2+L+w)w + (2+L+w)L = 100.5

(2+w -1-w/2 +/1/2w)sqr(w^4 +4w^3 +4w^2 +1764) +(2+w -1-w/2+/-(1/2)sqr(w^4 +4w^3 +4w^2 +1764)(-1-w/2 +/- (1/2w)sqr(w^4 +4w^3 +4w^2 + 1764)

now you reduced it to one equation one unknown, solve for w= width, graph it find the x intercepts if any, x intercepts = width, but look out for possible extraneous solutions

width = about 11.31

Length = about - 0.001

height = 2+ 11.31-.001 = 13.3099 but that makes no sense

there may be a problem with the problem

or

maybe the height = 2 more than both width and length meant h = 2+w and h =2+L, so w=L

that reduces the problem to 2 unknowns, making solving it far easier

let x = w=L then

V = 441 = hx^2, h = 441/x^2

A = 301 = 2hx +2hx +2x^2 = 4hx +2x^2

x^2 +2hx -100.5 = 0

x^2 +2(441/x^2)x - 100.5

x^2 +882/x = 100.5

x^3 -100.5x + 882 = 0 use the cubic formula or graph it

x = just one negative solution and two imaginary solutions

odds are the problem is with the problem, a typo, mistake in copying or a misprint