Quadratic Optimization: Additional Practice (Set 1)

With one number $x$, the other is $24 - x$, so the product is $P = x(24 - x) = -x^2 + 24x$, largest at the vertex.

$x = -\frac{24}{2(-1)} = 12 \;\Rightarrow\; P = 12 \cdot 12 = 144$

The greatest product is $144$, from the equal pair $12$ and $12$.

Expanded, $R = -3p^2 + 150p$, so the best price is the vertex input with $a = -3$, $b = 150$.

$p = -\frac{150}{2(-3)} = 25$

A price of $25$ dollars maximizes revenue.

Since $a = 1 > 0$ the cost has a minimum at $x = -\tfrac{-14}{2} = 7$. Substitute back for the cost.

$C = (7)^2 - 14(7) + 60 = 49 - 98 + 60 = 11$

The least cost is $11$ dollars, reached at the setting $x = 7$.

With width $x$ the length is $80 - x$ (half the perimeter minus the width), so $A = x(80 - x) = 80x - x^2$. Since $a = -1 < 0$ it opens downward, so the maximum is at the vertex.

$x = -\frac{80}{2(-1)} = 40, \qquad A = 40(80 - 40) = 40 \cdot 40 = 1600$

The greatest area is $1600$ square metres, from a $40$ by $40$ square.

Let $x$ be the number of $1$ dollar increases. The fee is $10 + x$ and the subscribers number $80 - 5x$, so $R = (10 + x)(80 - 5x) = -5x^2 + 30x + 800$.

$x = -\frac{30}{2(-5)} = 3 \;\Rightarrow\; R = (13)(65) = 845$

The maximum monthly revenue is $845$ dollars, at a fee of $13$ dollars with $65$ subscribers.

Let $x$ be each side out from the wall; the area is $A = x(120 - 2x)$, maximized at $x = -\tfrac{120}{2(-2)} = 30$.

$\text{parallel side} = 120 - 2(30) = 60$

The side parallel to the wall is $60$ metres. The value $1800$ is the area, not a side length.

Expanded, $R = -x^2 + 90x$, largest at the price $x = -\tfrac{90}{2(-1)} = 45$. Substitute back.

$R = 45(90 - 45) = 45 \cdot 45 = 2025$

The greatest revenue is $2025$ dollars, at a price of $45$ dollars.

The peak is at $t = -\tfrac{80}{2(-16)} = 2.5$ seconds. Substitute back for the height.

$h = -16(2.5)^2 + 80(2.5) + 10 = -100 + 200 + 10 = 110$

The maximum height is $110$ feet. The value $100$ feet leaves out the $10$ foot platform, and $2.5$ seconds is the time.

With width $x$ the length is $18 - x$, so $A = x(18 - x) = 18x - x^2$. Since $a = -1 < 0$ it opens downward, so the maximum is at the vertex.

$x = -\frac{18}{2(-1)} = 9, \qquad A = 9(18 - 9) = 9 \cdot 9 = 81$

The greatest area is $81$, from a $9$ by $9$ square.

Let $x$ be the number of $5$ dollar increases. The rate is $60 + 5x$ and the rooms filled are $80 - 5x$, so $R = (60 + 5x)(80 - 5x) = -25x^2 + 100x + 4800$.

$x = -\frac{100}{2(-25)} = 2 \;\Rightarrow\; \text{rate} = 60 + 5(2) = 70$

A rate of $70$ dollars maximizes revenue (renting $70$ rooms).

Let $x$ be each side out from the wall; the parallel side is $300 - 2x$, so $A = x(300 - 2x) = -2x^2 + 300x$.

$x = -\frac{300}{2(-2)} = 75 \;\Rightarrow\; A = 75(300 - 150) = 75 \cdot 150 = 11250$

The greatest area is $11250$ square feet, from a $75$ by $150$ pen.

Expanded, $R = -2p^2 + 200p$, largest at the price $p = -\tfrac{200}{2(-2)} = 50$. Substitute back.

$R = 50(200 - 2 \cdot 50) = 50 \cdot 100 = 5000$

The greatest revenue is $5000$ dollars, at a price of $50$ dollars.