how do i do part d (1 Viewer)

5uckerberg · Nov 23, 2021

Q10a) Surface area is the area from the outside and what does a box look like then solve. Get rid of the lid which is just x squared. 2x squared minus x squared.
b) Do you know how to find the volume of a box? If so then a piece of cake. After that sub h into the previous part.
c) Differentiate and find S'(x)=0 and then sub that value into the original qiuestion.

CM_Tutor · Nov 23, 2021

$\begin{align*} S &= A_{\text{base}} + A_{\text{sides}} \\ &= A_{\text{base}} + 4 \times A_{\text{side}} \qquad \text{since all 4 sides are identical} \\ &= x^2 + 4xh \end{align*}$

$\begin{align*} \text{Since}\ V &= A_{\text{base}} \times h \\ 32 &= x^2 h \\ h &= \cfrac{32}{x^2} \\ \text{Putting this into} \qquad S &= x^2 + 4xh \\ &= x^2 + 4x \times \cfrac{32}{x^2} \\ &= x^2 + \cfrac{128}{x} \end{align*}$

$\begin{align*} \text{Since}\ S &= x^2 + \cfrac{128}{x} \\ \cfrac{dS}{dx} &= 2x + 128 \times \cfrac{-1}{x^2} \\ &= 2x - \cfrac{128}{x^2} \\ \text{Put $\cfrac{dS}{dx} = 0$:} \qquad 2x - \cfrac{128}{x^2} &= 0 \\ 2x^3 &= 128 \\ x &= 4 \\ \text{Now,}\ \cfrac{d^2S}{dx^2} &= 2 - 128 \times \cfrac{-2}{x^3} \\ &= 2 + \cfrac{256}{x^3} \\ \text{At $x = 4$:} \qquad \cfrac{d^2S}{dx^2} &= 2 + \cfrac{256}{4^3} \\ &= 6 > 0\ \implies\ S\ \text{has a minimum when $x = 4$} \end{align*}$

So, the box with minimum surface area has a base of 4 m x 4 m and a height of 2 m, and the minimised surface area is 48 m².

how do i do part d (1 Viewer)

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