If the minimum value of y = 1 + hx - 3x\(^2\) is 13, find h.
13
12
11
10
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If the
minimum value of y = 1
+ hx - 3x^2 is 13, find h.
solution
y = 1 + hx - 3x^2
dy/dx = h - 6x
h - 6x = 0 (at turning point)
then, x = h/6
therefore: 13 = 1+ h.h/6 - 3.(h/6)^2
12 = h^2/6 - 3.h^2/36
12 = h^2/ - h^2/12
==> h^2 = 144
h = sq. root of 144
then, h is 12
hence d ans. is 12. NOT 11 TAKE NOTE PLEASE, I ALSO MAKE MISTAKE. THANKS.
y= 1+hx-3x^2
the minimum of value y, must exist wen there is a minimum value of x. .
To get the values or value of x,we need to differentiate d equation with respect to x.
dy/dx = h- 6x
at stationary points, dy/dy = 0
h - 6x = 0
subtract both sides by 6x , we HV ;
h = 6x
divide both side by 6;
x = h/ 6
substitute h/6 for x in d equation; and 13 for y.
y=1+hx-3x^2
13= 1 + h(h/6) -3(h/6)^2
13= 1+ h^2/6 - h^2/12
multiply both sides by 12.
156= 12 + 2h^2 - h^2
156= 12 + h^2
subtracting 12 from both sides,we HV;
144= h^2
square root both sides;.
h= 12
just substitute 13 for H in the equation, then you will have y=1+13(x) - 3x. Y= 11 is the answer
Thanks for your contributions. Corrections have been made, the correct answer is B.
y =1 + hx - 3x^2
13 = 1 + hx -3x^2
therefore
h = 14 - 3x^2/x
h = 11, since x cancelled x above. then h = 11
