53 - 55 Solved Problems in Maxima and Minima

$(0.5y)^2 + (0.5y + x)^2 = a^2$

$(0.5y + x)^2 = a^2 - 0.25y^2$

$x = \sqrt{a^2 - 0.25y^2} - 0.5y$
 

Area of rectangle
$A = xy$

$A = \left( \sqrt{a^2 - 0.25y^2} - 0.5y \right) \, y$

$A = y \sqrt{a^2 - 0.25y^2} - 0.5y^2$

$\dfrac{dA}{dy} = y \dfrac{-0.5y}{2\sqrt{a^2 - 0.25y^2}} + \sqrt{a^2 - 0.25y^2} - y = 0$

$\dfrac{-0.5y^2 + 2(a^2 - 0.25y^2) - 2y\sqrt{a^2 - 0.25y^2}}{2\sqrt{a^2 - 0.25y^2}} = 0$

$-y^2 + 4(a^2 - 0.25y^2) - 4y\sqrt{a^2 - 0.25y^2} = 0$

$4a^2 - 2y^2 = 4y\sqrt{a^2 - 0.25y^2}$

$2a^2 - y^2 = 2y\sqrt{a^2 - 0.25y^2}$

$4a^4 - 4a^2 \, y^2 + y^4 = 4y^2 \, (a^2 - 0.25y^2)$

$2y^4 - 8a^2 \, y^2 + 4a^4 = 0$

$y^4 - 4a^2 \, y^2 + 2a^4 = 0$

$y^2 = \dfrac{4a^2 \pm \sqrt{16a^4 - 8a^4}}{2}$

$y^2 = \dfrac{4a^2 \pm 2.828a^2}{2}$

$y^2 = 3.414a^2 \, \text{ and } \, 0.586a^2$
 

for
$y^2 = 3.414a^2$

$y = 1.848a$
 

$x = \sqrt{a^2 - 0.25(3.414a^2)} - 0.5(1.848a)$

$x = -0.541a$       (meaningless)
 

for
$y^2 = 0.586a^2$

$y = 0.765a$
 

$x = \sqrt{a^2 - 0.25(0.586a^2)} - 0.5(0.765a)$

$x = 0.541a$           answer

Let
k = cost per unit area of tin
mk = cost per unit area of copper
C = total cost
 

$C = (\pi r^2)mk + (2\pi rh)k$

$C = mk\pi r^2 + 2k\pi rh$

$\dfrac{dC}{dr} = 2mk\pi r + 2k\pi \left( r \dfrac{dh}{dr} + h \right) = 0$

$mr + r \dfrac{dh}{dr} + h = 0$

$\dfrac{dh}{dr} = -\dfrac{mr + h}{r}$

Volume
$V = \pi r^2 \, h$

$\dfrac{dV}{dr} = \pi \left( r^2 \dfrac{dh}{dr} + 2rh \right) = 0$

$\dfrac{dh}{dr} = -\dfrac{2h}{r}$
 

$\dfrac{dh}{dr} = \dfrac{dh}{dr}$

$-\dfrac{mr + h}{r} = -\dfrac{2h}{r}$

$mr + h = 2h$

$h = mr$

Height = m × radius           answer

$C = (\pi r^2)mk + (2\pi rh)k + (\pi r^2)k$

$C = k\pi(m + 1)r^2 + 2k\pi rh$

$\dfrac{dC}{dr} = 2k\pi (m + 1)r + 2k\pi \left( r \dfrac{dh}{dr} + h \right) = 0$

$(m + 1)r + r \dfrac{dh}{dr} + h = 0$

$\dfrac{dh}{dr} = -\dfrac{(m + 1)r + h}{r}$
 

Volume
$V = \pi r^2 \, h$

$\dfrac{dV}{dr} = \pi \left( r^2 \dfrac{dh}{dr} + 2rh \right) = 0$

$\dfrac{dh}{dr} = -\dfrac{2h}{r}$
 

$\dfrac{dh}{dr} = \dfrac{dh}{dr}$

$-\dfrac{(m + 1)r + h}{r} = -\dfrac{2h}{r}$

$(m + 1)r + h = 2h$

$h = (m + 1)r$

Height = (m + 1) × radius           answer