Asked by Cynthia
an open box contains 80cm^3 and is made from a square piece of tinplate with 3cm squares cut from each of its 4 corners. Find the dimensions of the original piece of tinplate.
Answers
Answered by
MathMate
Let the side of the square plate be x cm.
The height of the box is 3 cm.
The length & width are both (x-6) cm.
volume = length*width*height
=3*(x-6)*(x-6)
=80
x can then be obtained by solving
3(x-6)(x-6)=80
The height of the box is 3 cm.
The length & width are both (x-6) cm.
volume = length*width*height
=3*(x-6)*(x-6)
=80
x can then be obtained by solving
3(x-6)(x-6)=80
Answered by
Reiny
I see the equation
3(x-6)(x-6) = 80
3(x^2 - 12x + 36) = 80
3x^2 - 36x + 108 - 80 = 0
3x^2 - 36x + 28 = 0
Use the quadratice formula to find x
I got x=11.164
check 11.164-6 = 5.164
so the box is 5.164x5.164x3
which gives 80
3(x-6)(x-6) = 80
3(x^2 - 12x + 36) = 80
3x^2 - 36x + 108 - 80 = 0
3x^2 - 36x + 28 = 0
Use the quadratice formula to find x
I got x=11.164
check 11.164-6 = 5.164
so the box is 5.164x5.164x3
which gives 80
Answered by
Evan
So why is the length and width x-6? where did the 6 come from can someone explain please?
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