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A rectangular piece of tin has an area of 1056 square inches. A square of 3 inches is cut from each corner, and an open box is made by turning up the ends and sides. If the volume of the box is 2052 cubic inches, what were the original dimensions of the piece of tin. PLEASE SHOW WORK

Sagot :

Answer:

44 inches and 24 inches

Step-by-step explanation:

Let [tex]l,b[/tex] denotes length and breadth of a rectangle.

[tex]lb=1056[/tex]

A square of 3 inches is cut from each corner, and an open box is made by turning up the ends and sides.

New length [tex]=l-6[/tex] inches

New breadth [tex]=b-6[/tex] inches

Height of box [tex]=3[/tex] inches

Volume of box = length × breadth × height

[tex](l-6)(b-6)3=2052[/tex]

[tex]lb-6l-6b+36=684\\6(l+b)=408\\l+b=68\\b=68-l[/tex]

Put [tex]b=68-l[/tex] in [tex]lb=1056[/tex]

[tex]l(68-l)=1056\\l^2-68l+1056=0\\l^2-44l-24l+1056=0\\l(l-44)-24(l-44)=0\\(l-24)(l-44)=0\\l=24,44[/tex]

At [tex]l=24,b=\frac{1056}{24}=44[/tex] inches

At [tex]l=44,b=\frac{1056}{44}=24[/tex] inches