A net of an open box is shown. The box is a cuboid with height $h$ centimetres. The base is a rectangle measuring $3x$ centimetres by $2x$ centimetres. (a) (i) Express the area of the net, $A ext{ cm}^2$, in terms of $h$ and $x$. (ii) Given that $A = 720 ext{ cm}^2$, show that the volume of the box, $V ext{ cm}^3$, is given by $V = rac{4320x - rac{18}{5}x^3}{3}$. (b) Determine the value of $x$ that maximises the volume of the box.

Scottish Highers Maths Differentiation: A net of an open box is shown. T

A net of an open box is shown - Scottish Highers Maths - Question 14 - 2023

Question 14

A net of an open box is shown. The box is a cuboid with height $h$ centimetres. The base is a rectangle measuring $3x$ centimetres by $2x$ centimetres. (a) (i) Expr... show full transcript

Worked Solution & Example Answer:A net of an open box is shown - Scottish Highers Maths - Question 14 - 2023

Step 1

Express the area of the net, $A ext{ cm}^2$, in terms of $h$ and $x$.

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Answer

The area of the net of the open box can be calculated by adding the areas of all the faces. The base has an area of $6x^2$ ( $3x imes 2x$ ) and the four sides contribute an additional area of $2(3x imes h) + 2(2x imes h)$ , simplifying to $10xh$ . Therefore, the total area $A$ is given by:

$A = 6x^2 + 10xh$

Step 2

Given that $A = 720 ext{ cm}^2$, show that the volume of the box, $V ext{ cm}^3$, is given by $V = \frac{4320x - \frac{18}{5}x^3}{3}$.

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Answer

Setting $A = 720$ leads us to:

$720 = 6x^2 + 10xh$

Rearranging yields:

$10xh = 720 - 6x^2 \\ h = \frac{720 - 6x^2}{10x}$

The volume $V$ of the cuboid is given by:

$V = \text{Base Area} \times h = (6x^2) \times h = 6x^2 \times \frac{720 - 6x^2}{10x}$

Simplifying this gives:

$V = 4320 - \frac{36}{5}x^2$

Which can be rewritten as:

$V = \frac{4320x - \frac{18}{5}x^3}{3}$

Step 3

Determine the value of $x$ that maximises the volume of the box.

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Answer

To find the maximum volume, we differentiate $V$ with respect to $x$ :

$V'(x) = 4320 - \frac{54}{5}x^2$

Setting this derivative to zero to find critical points:

$0 = 4320 - \frac{54}{5}x^2 \\ \frac{54}{5}x^2 = 4320 \\ x^2 = \frac{4320 \times 5}{54} = 400 \\ x = 20$

To confirm that this gives a maximum, we differentiate $V'(x)$ :

$V''(x) = -\frac{108}{5}x$

At $x = 20$ :

$V''(20) = -\frac{108}{5}(20) < 0$

Thus, $x = 20$ maximizes the volume.

A net of an open box is shown - Scottish Highers Maths - Question 14 - 2023

Worked Solution & Example Answer:A net of an open box is shown - Scottish Highers Maths - Question 14 - 2023

Express the area of the net, $A ext{ cm}^2$, in terms of $h$ and $x$.

Given that $A = 720 ext{ cm}^2$, show that the volume of the box, $V ext{ cm}^3$, is given by $V = \frac{4320x - \frac{18}{5}x^3}{3}$.

Determine the value of $x$ that maximises the volume of the box.

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