The Volume of a tin of radius r cm is given by V=pi*(40r-r^2-r^3). Find the positive value of r for which dV/dr=0 and find the value of V for this r.

Firstly differentiate the function V with respect to x (dV/dx)=pi*(40-2r-3r^2). Set dV/dr =0 and solve to find r. Divide both sides by pi and divide both side by -1 so that the r^2 term is positive (I personally find it easier to solve when the highest power coefficient is positive) (3r^2+2r-40=0). Factorise this expression ((3r-10)(r+4)=0). From this you can deduce the two values of x for which dV/dx=0 (r=10/3 and r =-4). Lastly substitute the positive value of r (10/3) into the expression for V (V=pi*(40-2(10/3)-3(10/3)^2)) = (2300*pi)/27.

CK

Related Maths A Level answers

All answers ▸

Solve for x, between 0 and 360 degrees, 4cos2 (x) + 7sin (x) – 2 = 0


A curve has the equation: x^3 - x - y^3 - 20 = 0. Find dy/dx in terms of x and y.


Functions: If f(x)=3x^2 - 4 and g(x) = x + 3, 1) Evaluate f(3), 2) Find the inverse of f(x) (f^-1(x)), 3)Find fg(x).


The function f has domain (-∞, 0) and is defines as f(x) = (x^2 + 2)/(x^2 + 5) (here ^ is used to represent a power). Show that f'(x) < 0. What is the range of f?