A cuboid has a square base of sides \(x\text{ cm}\). Its height is \(h\text{ cm}\) and its volume is \(150\text{ cm}^3\).
Show that the total surface area of the cuboid, \(A=2x^2+\dfrac{600}{x}\).[2]
Complete the table of values for \(A=2x^2+\dfrac{600}{x}\).[1]
| \(x\) | \(2\) | \(4\) | \(6\) | \(8\) | \(10\) | \(12\) | \(14\) | \(16\) | \(18\) | \(20\) |
|---|
| \(A\) | \(308\) | \(182\) | \(172\) | \(203\) | \(260\) | \(338\) | | \(550\) | \(681\) | \(830\) |
On the grid, draw the graph of \(A=2x^2+\dfrac{600}{x}\) for \(2\leq x\leq20\).[3]
Can the total surface area of the cuboid be \(150\text{ cm}^2\)? Explain your answer.[1]
A manufacturer designs cuboid packaging boxes with square bases. The total surface area must not exceed the material cost limit which is given by \(A=25x+150\).
On the same grid, draw the straight line \(A=25x+150\) for \(2\leq x\leq20\).[1]
Use the graph to find the possible lengths of the sides of the cuboid where the material cost is maximised.[2]