Question 985067
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L=length=50-2x; W=width=20-2x; H=height=x; V=volume
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{{{V=LWH}}}
{{{V=(50-2x)(20-2x)(x)}}}
{{{V=4x^3-140x^2+1000x}}}
The function will have a maximum at the point where the first derivative equals zero.
{{{dV/dx=12x^2-280x+1000}}}
{{{12x^2-280x+1000=0}}}
{{{3x^2-70x+250=0}}}
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*[invoke quadratic "x", 3, -70, 250 ]
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Determine the domain of the function. x must be greater than 0 or H would be zero and we would have a flat sheet, so x>0. 2x must be less than 20 or we would have no width so 2x<20 or x<10
So 0 < x < 10.
That leaves us with x=4.40 as the solution.
The size of the box:
L=50-2x=50-8.80=41.20
W=20-2x=20-8.80=11.20
H=x=4.40
Maximum Volume of the box=(41.20)(11.20)(4.40)=2030.34 cm^3