A company has to design a rectangular box for a new range of jellybeans - Leaving Cert Mathematics - Question 7 - 2014

The capacity of the box $V$ can be expressed as:

$V = ext{length} imes ext{width} imes ext{height}$

Substituting the expressions for length, width, and height:

$V = (15 - h)(20 - 2h)(h)$

Expanding this gives:

= 2h^3 - 50h^2 + 300h \, cm³$$

For the bottom of the box to be square, the length must equal the width:

Setting length equal to width:

$15 - h = 20 - 2h$

Solving this equation:

h = 5 $$ Using $h = 5$ into the capacity equation $$ V = (15 - 5)(20 - 2 imes 5)(5) \ = (10)(10)(5) \ \\ = 500 \, cm³$$ Thus, $h = 5$ gives the correct capacity.

To find the other value of $h$, we must solve the equation for capacity:

$2h^3 - 50h^2 + 300h - 500 = 0$

Factoring out the previous value of $h = 5$ provides:

$2h^3 - 50h^2 + 300h - 500 = (h - 5)(2h^2 - 40h + 100)$

Now, using the quadratic formula on $2h^2 - 40h + 100 = 0$ leads to:

= \frac{40 \pm \sqrt{1600 - 800}}{4} \ = \frac{40 \pm \sqrt{800}}{4} \ = \frac{40 \pm 20\sqrt{2}}{4} \ = 10 \pm 5\sqrt{2}$$ Calculating both results gives: $$h \approx 17.1, 2.9$$ The other value of $h$ is approximately 2.9 when rounded to one decimal place.

The client requires a new box with a capacity increased by 10%:

$550 = 1.1 \times 500$

Referring to the graph, the capacity represented will only intersect with the horizontal line at one point, which is greater than 15 for $h$. The maximum allowable height is set to ensure that $h$ must be greater than 1 and less than 15 for feasibility. Thus, if the height exceeds 15, it violates the material constraints of the original cardboard.

Therefore, it is not physically possible to construct a larger box from the same piece of cardboard.