'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h). Die lengte (l) van die basis moet 3 keer die breedte (w) wees. Die volume moet 5 m³ wees. Die materiaal vir die bokant en onderkant kos R15 per vierkante meter en die materiaal vir die sykante kos R6 per vierkante meter. 9.1 Toon dat die koste om die houer te vervaardig soos volg bereken kan word: Koste = 90w² + 48wh 9.2 Bepaal die breedte van die houer sodat die koste om die houer te vervaardig, in minimum sal wees.

NSC Mathematics Calculus: 'n Toe rehogike houder (boks) mo

'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h) - NSC Mathematics - Question 9 - 2020 - Paper 1

Question 9

'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h). Die lengte (l) van die basis moet 3 keer die breedte... show full transcript

Worked Solution & Example Answer:'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h) - NSC Mathematics - Question 9 - 2020 - Paper 1

Step 1

9.1 Toon dat die koste om die houer te vervaardig soos volg bereken kan word:

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Answer

To solve for the cost of manufacturing the box, we first need to determine the relationship between the dimensions given:

We know that the length (l) is three times the width (w):
$l = 3w$
The volume (V) of the box is given by the formula:
$V = l \times w \times h$ Given that the volume must be 5 m³:
$5 = (3w) \times w \times h$
Therefore,
$h = \frac{5}{3w^2}$
The surface area (SA) of the box, which includes top, bottom, and sides, can be calculated as follows:
$SA = 2lw + 2lh + 2wh$ Substituting for l and h, we have:
$SA = 2(3w)w + 2(3w)\left(\frac{5}{3w^2}\right) + 2w\left(\frac{5}{3w^2}\right)$
By simplifying, we arrive at the cost function:
$C = 90w^2 + 48h = 90w^2 + 48\left(\frac{5}{3w^2}\right)$
This leads to the formulation
$C = 90w^2 + 80w^{-1}$ .

Step 2

9.2 Bepaal die breedte van die houer sodat die koste om die houer te vervaardig, in minimum sal wees.

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Answer

To find the minimum cost, we need to differentiate the cost function with respect to w. From the previous part, we have:

The cost function:
$C(w) = 90w^2 + 48\left(\frac{5}{3w^2}\right)$
To find the derivative of C with respect to w, we apply the power rule:
$C'(w) = 90 \cdot 2w - 48 \cdot \left(\frac{5 \cdot 2}{3w^3}\right)$
This simplifies to:
$C'(w) = 180w - \frac{480}{3w^3}$
Setting the derivative equal to zero allows us to find critical points:
$180w - 160w^{-3} = 0$
Rearranging gives:
$180w^4 = 160$
Thus:
$w^4 = \frac{16}{18} = \frac{8}{9}$
Therefore,
$w = \sqrt[4]{\frac{8}{9}} = \frac{2}{3\sqrt{3}} \approx 0.76 \text{ m}$
Hence, the width w that minimizes the cost is approximately 0.76 m.

'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h) - NSC Mathematics - Question 9 - 2020 - Paper 1

Worked Solution & Example Answer:'n Toe rehogike houder (boks) moet soos volg vervaardig word: Afmetings: lengte (l), breedte (w) en hoogte (h) - NSC Mathematics - Question 9 - 2020 - Paper 1

9.1 Toon dat die koste om die houer te vervaardig soos volg bereken kan word:

9.2 Bepaal die breedte van die houer sodat die koste om die houer te vervaardig, in minimum sal wees.

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