1. **Stating the problem:** We have a square sheet of cardboard with side length 10 units. Squares of side length $x$ units are cut from each corner, and the sheet is folded to form an open box. We need to express the box dimensions in terms of $x$, find the volume formula, explain why a box of height 6 units cannot be made, complete a table of volumes for given $x$ values, and estimate values from the graph. 2. **Expressing dimensions:** - Length $l$ is the original length minus two cut squares: $$l = 10 - 2x$$ - Width $b$ is the same as length since the sheet is square: $$b = 10 - 2x$$ - Height $h$ is the side length of the cut squares: $$h = x$$ 3. **Volume formula:** Volume $V$ is length times width times height: $$V = l \times b \times h = (10 - 2x)(10 - 2x)(x) = (10 - 2x)^2 x$$ 4. **Expanding volume:** First expand $(10 - 2x)^2$: $$ (10 - 2x)^2 = 100 - 40x + 4x^2 $$ Multiply by $x$: $$ V = x(100 - 40x + 4x^2) = 100x - 40x^2 + 4x^3 $$ Rearranged: $$ V(x) = 4x^3 - 40x^2 + 100x $$ 5. **Why height 6 units is impossible:** Height $h = x$. For $h=6$, $x=6$. But since $l = 10 - 2x$, substituting $x=6$ gives: $$ l = 10 - 2(6) = 10 - 12 = -2 $$ Length cannot be negative, so $x=6$ is not possible. Hence, a box with height 6 units cannot be made. 6. **Completing the table:** Calculate $V(x)$ for given $x$ values: - $V(0) = 4(0)^3 - 40(0)^2 + 100(0) = 0$ - $V(0.5) = 4(0.5)^3 - 40(0.5)^2 + 100(0.5) = 4(0.125) - 40(0.25) + 50 = 0.5 - 10 + 50 = 40.5$ - $V(1) = 4(1) - 40(1) + 100(1) = 4 - 40 + 100 = 64$ - $V(1.5) = 4(3.375) - 40(2.25) + 100(1.5) = 13.5 - 90 + 150 = 73.5$ - $V(2) = 4(8) - 40(4) + 100(2) = 32 - 160 + 200 = 72$ - $V(2.5) = 4(15.625) - 40(6.25) + 100(2.5) = 62.5 - 250 + 250 = 62.5$ - $V(3) = 4(27) - 40(9) + 100(3) = 108 - 360 + 300 = 48$ - $V(3.5) = 4(42.875) - 40(12.25) + 100(3.5) = 171.5 - 490 + 350 = 31.5$ - $V(4) = 4(64) - 40(16) + 100(4) = 256 - 640 + 400 = 16$ - $V(4.5) = 4(91.125) - 40(20.25) + 100(4.5) = 364.5 - 810 + 450 = 4.5$ - $V(5) = 4(125) - 40(25) + 100(5) = 500 - 1000 + 500 = 0$ 7. **Graph and estimation:** (i) To find the maximum volume, observe the graph or calculate derivative and find critical points. The maximum volume is approximately $73.5$ units³ at $x=1.5$. (ii) To find $x$ values for volume 30 units³, solve: $$4x^3 - 40x^2 + 100x = 30$$ This cubic can be solved graphically or numerically. Approximate solutions are near $x=0.4$ and $x=3.8$. (iii) For $l = 2.8$ units, solve: $$10 - 2x = 2.8 \implies 2x = 7.2 \implies x = 3.6$$ Calculate volume at $x=3.6$: $$V(3.6) = 4(3.6)^3 - 40(3.6)^2 + 100(3.6) = 4(46.656) - 40(12.96) + 360 = 186.624 - 518.4 + 360 = 28.224$$ Final answers: - $l = 10 - 2x$ - $b = 10 - 2x$ - $h = x$ - $V(x) = 4x^3 - 40x^2 + 100x$ - Height 6 units impossible because $l$ becomes negative - Table values as above - Maximum volume approx 73.5 at $x=1.5$ - $x$ for volume 30 approx 0.4 and 3.8 - Volume at $l=2.8$ (i.e. $x=3.6$) approx 28.224