Question 5
Problem:
A solid cuboid has width $x \, \mathrm{cm}$, length $4x \, \mathrm{cm}$, and height $h \, \mathrm{cm}$. The volume is $75 \, \mathrm{cm}^3$, and the surface area is $S \, \mathrm{cm}^2$.
- (a) Show that $ S = 8x^2 + \frac{375}{2x} $.
- (b)
- (i) Find, to 3 significant figures, the value of $x$ for which $S$ is a minimum.
- (ii) Justify that this value gives a minimum.
- (c) Find, to 3 significant figures, the minimum value of $S$.
Solution:
(a)
Volume:
$ 4x^2 h = 75 \Rightarrow h = \frac{75}{4x^2} $
Surface area:
$ S = 2(4x^2 + xh + 4xh) = 2(4x^2 + 5xh) $
Substitute $h$:
$ S = 2\left(4x^2 + \frac{375}{4x}\right) = 8x^2 + \frac{375}{2x} $
(b)(i)
$ \frac{dS}{dx} = 16x - \frac{375}{2x^2} $
Set to zero:
$ 16x - \frac{375}{2x^2} = 0 $
$ 32x^3 - 375 = 0 \Rightarrow x^3 = \frac{375}{32} $
$ x = \sqrt[3]{\frac{375}{32}} \approx 2.27 \, \mathrm{cm} $
(b)(ii)
$ \frac{d^2S}{dx^2} = 16 + \frac{375}{x^3} $
Since this is positive, $S$ is minimized.
(c)
$ S = 8x^2 + \frac{375}{2x} $
Substitute $x \approx 2.27$:
$ S \approx 124 \, \mathrm{cm}^2 $
Final Answer: $124 \, \mathrm{cm}^2$
Question 6
Problem:
Solve the equation
$ \log_2 x^3 + \log_4 x^2 - 3 \log_x 2 = 0 $
giving your answers to 3 significant figures.
Solution:
Step 1:
$ \log_2 x^3 = 3\log_2 x $
Step 2:
$ \log_4 x^2 = 2\log_4 x = 2 \cdot \frac{\log_2 x}{2} = \log_2 x $
Step 3:
$ \log_x 2 = \frac{1}{\log_2 x} \quad \Rightarrow \quad -3\log_x 2 = -\frac{3}{\log_2 x} $
Step 4:
$ 3\log_2 x + \log_2 x - \frac{3}{\log_2 x} = 0 $
$ 4\log_2 x - \frac{3}{\log_2 x} = 0 $
Step 5:
$ 4(\log_2 x)^2 - 3 = 0 $
$ (\log_2 x)^2 = \frac{3}{4} $
$ \log_2 x = \pm \frac{\sqrt{3}}{2} $
Step 6:
$ x = 2^{\frac{\sqrt{3}}{2}} \quad \text{or} \quad x = 2^{-\frac{\sqrt{3}}{2}} $
Step 7:
$ x \approx 1.82 \quad \text{or} \quad x \approx 0.549 $
Final Answer:
$ \boxed{x \approx 1.82 \quad \text{and} \quad x \approx 0.549} $
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