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Please answer this question below, please

Please answer this question below, please-example-1
User EngJon
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so... if we notice the picture, the height of the finished box is 3 then

the length is "8 longer than the width" or w + 8 then

thus
\bf \textit{volume of a rectangular prism}\\\\ V=lwh\qquad \begin{cases} l=length\\ w=width\\ h=height\\ -------\\ l=w+8\\ h=3\\ V=315 \end{cases}\implies 315=(w+8)(w)(3) \\\\\\ \cfrac{315}{3}=w^2+8w\implies 0=w^2+8w-105 \\\\\\ 0=(w+15)(w-7)

well, -15 is clearly not a feasible value for a width... so the value is 7 then

now... that's just the finished box... notice the picture, the width on the cardboard is really 3 + width + 3, and the length is 3 + length + 3,
so you're really adding 6 or 3+3, to each, so in this case, the width is 7+6

and the length...well, you'd know that already by now
Please answer this question below, please-example-1
User Kardux
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