Question

please complete both question
this is urgent so please try and complete it as fast as possible
due today ​

Answer 1

Answer:

$\textsf{a)} \quad x=\dfrac{25(7-\sqrt{13})}{3}=28.2870727\; \sf mm$

$\textsf{b)} \quad V=379037.8082\; \sf mm^3$

Explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Volume of a rectangular prism}\\\\ V=w\:l\:h \\\\where:\\ \phantom{ww} \bullet w is the width of the base. \\ \phantom{ww} \bullet l is the length of the base. \\ \phantom{ww} \bullet h is the height.\\\end{minipage}}$

Given:

• l = 200 mm
• w = 150 mm

The dimensions of the rectangular prism in terms of x are:

• Length = 200 - 2x
• Width = 150 - 2x
• Height = x

Substitute these values into the formula for volume to create an equation in terms of x:

$\begin{aligned}\implies V&=(150-2x)(200-2x)x\\&=(30000-700x+4x^2)x\\&=4x^3-700x^2+30000x\end{aligned}$

To find the value of x that will give the maximum volume, differentiate the equation for volume.

$\begin{aligned}\implies \dfrac{\text{d}V}{\text{d}x}&=3 \cdot 4x^{3-1}-2 \cdot 700x^{2-1}+30000x^{1-1}\\&=12x^2-1400x+30000\end{aligned}$

Set the derivative to zero and solve for x using the quadratic formula:

$\implies x=\dfrac{-(-1400) \pm \sqrt{(-1400)^2-4(12)(30000)}}{2(12)}$

$\implies x=\dfrac{1400 \pm \sqrt{520000}}{24}$

$\implies x=\dfrac{1400 \pm \sqrt{40000 \cdot 13}}{24}$

$\implies x=\dfrac{1400 \pm \sqrt{40000}\sqrt{13}}{24}$

$\implies x=\dfrac{1400 \pm 200\sqrt{13}}{24}$

$\implies x=\dfrac{175\pm 25\sqrt{13}}{3}$

$\implies x=\dfrac{25(7\pm\sqrt{13})}{3}$

To determine which value of x will give the maximum volume, differentiate again:

$\begin{aligned}\implies \dfrac{\text{d}^2V}{\text{d}x^2}&=2 \cdot12x^{2-1}-1400x^{1-1}+0\\&=24x-1400\end{aligned}$

Substitute both values of x into the second derivative:

$x=\dfrac{25(7+\sqrt{13})}{3} \implies \dfrac{\text{d}^2V}{\text{d}x^2}=721.1102551 > 0\implies \sf minimum$

$x=\dfrac{25(7-\sqrt{13})}{3} \implies \dfrac{\text{d}^2V}{\text{d}x^2}=-721.1102551 < 0\implies \sf maximum$

Therefore, the value of x which will give the maximum volume is:

$x=\dfrac{25(7-\sqrt{13})}{3}=28.2870727\; \sf mm$

To find the maximum volume of the box, substitute the found value of x into the equation for volume:

$\begin{aligned}\implies V_{\sf max}&=4\left(\dfrac{25(7-\sqrt{13})}{3}\right)^3-700 \left(\dfrac{25(7-\sqrt{13})}{3}\right)^2+30000 \left (\dfrac{25(7-\sqrt{13})}{3}\right)\\\\&=379037.8082\; \sf mm^3\end{aligned}$

The dimensions of the box with maximum possible volume will be:

• Length = 143.4 mm (1 d.p.)
• Width = 93.4 mm (1 d.p.)
• Height = 28.3 mm (1 d.p.)