
Larry C. answered 09/18/18
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Assuming that by 'cargo space of 10ft', you mean the cargo area is 10' deep, then the volume would be 10*w*(w-2)=10w2-20w So: 350=10w2-20w -> 35=w2-2w -> w2-2w-35=0 -> (w+5)(w-7)=0 which has solutions of -5 and 7. Since the width cannot be negative, the dimensions are 10x7x5