APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Find two positive integers whose sum is 16 and sum of whose cubes is minimum.
✓
Answer
Let the two positive numbers are x and y.
$\therefore\ \text{x}+\text{y}=16\ \Rightarrow\ \text{y}=16-\text{x}\ \dots\text{(i)}$
Let $\text{z}=\text{x}^3+\text{y}^3\ \Rightarrow \ \text{z}=\text{x}^3+(16-\text{x)}^3\ \ [\text{From eq.(i)}]$
$\Rightarrow\ \text{z}=\text{x}^3+(16)^3-\text{x}^2-48\text{x}(16-\text{x)}$ $=(16)^3-768\text{x}+48\text{x}^2$
$\Rightarrow\ \frac{\text{dz}}{\text{dx}}=-768+96\text{x} \ \text { and }\frac{\text{d}^2\text{z}}{\text{dx}^2}=96$
Now $ \frac{\text{dz}}{\text{dx}}=0\ \ \Rightarrow\ -768+96\text{x}=0\ \Rightarrow\ \text{x}=8$
$\text{At }\text{x}=8\ \frac{\text{d}^2\text{z}}{\text{dx}^2}=96 \ \text{ is positive}.$
$\therefore$ x = 8 is a point of local minima and z is minimum when x = 8.
$\therefore$ y = 16 - 8 = 8
Therefore, the required numbers are 8 and 8.
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