The dimensions of a box are b, b + 1, b + 4. Find how fast the total
surface area A increases as b increases.
1
Expert's answer
2020-12-14T18:29:27-0500
"\\text{Total Surface Area of the box(A)}=2(b(b+1)+(b+1)(b+4)+(b(b+4))\n\\\\=2((b^2+b)+(b^2+5b+4)+(b^2+4b))\\\\=2(3b^2+10b+4)\\implies 6b^2+20b+8\\\\\\text{Hence, } \\frac{dA}{db}=12b+20"
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