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tdabboud
The area of the paper is 260 in2. Then 2 inch square is cut from each corner and folded up to make a box. The volume of the box is 288 in3 what are the dimensions of the paper for this to be true? How do i do this?
Let x and y be the width and length of the uncut paper respectively. The area is given to be 260 sq inches, so\[x* y=260\]The length of the box width is x-2*2 because there are two each 2x2 inch squares cut out of each corner to make the box. The length of the box length is y-2*2 for the same reason. The box height will be 2 inches. The volume, 288 cubic inches, is equal to the product of the width*length*height or\[2(x-2*2)(y-2*2)= 288\]Solving the two simultaneous for x and y yields:\[\{x\to 13,y\to 20,x\to 20,y\to 13\} \]The uncut paper is 13 by 20 inches.
how did you solve them simultaneously?
x*y=260, y=260/x Replace y in the other equation with 260/x\[2(x-2*2)(y-2*2)=288\]\[2(x-2*2)\left(\frac{260}{x}-2*2\right)=288 \]\[552-\frac{2080}{x}-8 x-288=0\]Multiply each side by x and combine the resulting fractions.\[x\left(552-\frac{2080}{x}-8 x-288\right)= 0*x\]\[-8 \left(260-33 x+x^2\right)=0\]\[260 - 33 x + x^2 = 0 \]Use the binomial theorem or factor the LHS.\[(x-13)(x-20) =0 \]x=13 is the answer you want. Use y=260/x to find the corresponding y.