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Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 2
Question 2
Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal. The base of the tank is a rectangle x metres by y metres. The heigh... show full transcript
Worked Solution & Example Answer:Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 2 - 2007 - Paper 2
Step 1
Show that the area A m² of the sheet metal used to make the tank is given by A = \frac{300}{x} + 2x^2.
Answer
To find the area of the sheet metal used in the construction of the open-topped tank, we first note that the tank's volume is given by:
tank volume = length × width × height
Thus, we have:
[ V = x \cdot y \cdot x = x^2y ] Since the volume is 100 m³, we have:
[ x^2y = 100 \implies y = \frac{100}{x^2} ]
The surface area A of the tank is given by the sum of the areas of the four sides plus the base:
[ A = xy + 2xh + 2yh ] Substituting for y:
[ A = x \cdot \frac{100}{x^2} + 2x \cdot x + 2\left(\frac{100}{x^2}\right)x ] [ A = \frac{100}{x} + 2x^2 + \frac{200}{x} ] Thus,
[ A = \frac{300}{x} + 2x^2 ]
Step 2
Use calculus to find the value of x for which A is stationary.
Answer
To determine where A is stationary, we first differentiate A with respect to x:
[ \frac{dA}{dx} = -\frac{300}{x^2} + 4x ] Setting the derivative equal to zero gives:
[ -\frac{300}{x^2} + 4x = 0 ] Multiplying through by x² (to eliminate the fraction):
[ -300 + 4x^3 = 0 ] [ 4x^3 = 300 ] [ x^3 = 75 ] [ x = \sqrt[3]{75} \approx 4.22 ]
Step 3
Prove that this value of x gives a minimum value of A.
Answer
To confirm that x = \sqrt[3]{75} gives a minimum, we need to check the second derivative:
[ \frac{d^2A}{dx^2} = \frac{600}{x^3} + 4 ] Since both terms are positive for x > 0, we conclude that: [ \frac{d^2A}{dx^2} > 0 ] This indicates that A has a local minimum at x = \sqrt[3]{75}.
Step 4
Calculate the minimum area of sheet metal needed to make the tank.
Answer
Substituting x = \sqrt[3]{75} back into the area function:
[ A = \frac{300}{\sqrt[3]{75}} + 2(\sqrt[3]{75})^2 ]
To find A, calculate:
[ A \approx \frac{300}{4.22} + 2(17.44) \approx 71.05 + 34.88 \approx 105.93 \text{ m}^2 ]
Thus, the minimum area of sheet metal needed is approximately 105.93 m².