What is the extreme f(x)=((45-x)x-(10+24x)) ?

The extreme f(x)=((45-x)x-(10+24x)) is Maximum(21/2 , 401/4)

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extreme f(x)=((45-x)x-(10+24x))

extreme points

f (x) = ((45 - x) x - (10 + 24 x))

Maximum (\frac{21}{2}, \frac{401}{4})

Solution steps

f^{'} (x) = - 2 x + 21

Find intervals:

Increasing

: - \infty < x < \frac{21}{2},

Decreasing

: \frac{21}{2} < x < \infty

Plug

x = \frac{21}{2}

into

(45 - x) x - (10 + 24 x) : \frac{401}{4}

Maximum (\frac{21}{2}, \frac{401}{4})

y = 7 + 5 x - 5 x^{2}

T (x, y) = 46 - 2 x^{2} - 3 y^{2}

f (x) = (- x^{\frac{2}{3}}) (x - 3)

f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x) = f (x, y) = 3 x^{3} - 9 x + 9 x y^{2}

What is the extreme f(x)=((45-x)x-(10+24x)) ?
The extreme f(x)=((45-x)x-(10+24x)) is Maximum(21/2 , 401/4)