A variable straight-line L with negative slop | vedclass

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Question

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$,where $a, b > 0$.The equation of the line passing through $(a, 0)$ and $(0, b)$ is $\fr

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$,where $a, b > 0$.The equation of the line passing through $(a, 0)$ and $(0, b)$ is $\frac{x}{a} + \frac{y}{b} = 1$.Since the line passes through $(4, 9)$,we have $\frac{4}{a} + \frac{9}{b} = 1$.We want to minimize $S = a + b$.From the equation $\frac{4}{a} + \frac{9}{b} = 1$,we can express $b$ in terms of $a$: $\frac{9}{b} = 1 - \frac{4}{a} = \frac{a-4}{a}$,so $b = \frac{9a}{a-4}$.Thus,$S(a) = a + \frac{9a}{a-4}$.To find the minimum,we differentiate $S(a)$ with respect to $a$: $S'(a) = 1 + \frac{9(a-4) - 9a}{(a-4)^2} = 1 - \frac{36}{(a-4)^2}$.Setting $S'(a) = 0$,we get $(a-4)^2 = 36$,so $a-4 = 6$ (since $a > 4$),which gives $a = 10$.Then $b = \frac{9(10)}{10-4} = \frac{90}{6} = 15$.The minimum value is $S = a + b = 10 + 15 = 25$.

$A$ variable straight-line $L$ with negative slope passes through the point $(4,9)$ and cuts the positive coordinate axes in $A$ and $B$. If $O$ is the origin,then the minimum value of $OA+OB$ is

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