Quadratic Optimization: Additional Practice (Set 2)

Let $x$ be each side running out from the wall; the side parallel to the wall is $100 - 2x$, so the area is $A = x(100 - 2x) = -2x^2 + 100x$.

$x = -\frac{100}{2(-2)} = 25 \;\Rightarrow\; A = 25(100 - 50) = 25 \cdot 50 = 1250$

The greatest area is $1250$ square metres, from a $25$ by $50$ pen. The width $25$ is the input, not the area.

The peak is at $t = -\tfrac{64}{2(-16)} = 2$ seconds. Substitute back for the height.

$h = -16(2)^2 + 64(2) + 9 = -64 + 128 + 9 = 73$

The maximum height is $73$ feet. The time $2$ seconds answers "when", not "how high".

Name one number $x$; the other is $18 - x$, so the product is $P = x(18 - x) = -x^2 + 18x$, largest at the vertex.

$x = -\frac{18}{2(-1)} = 9 \;\Rightarrow\; P = 9(18 - 9) = 9 \cdot 9 = 81$

The two numbers are $9$ and $9$, and the largest product is $81$.

From $x + 3y = 60$ we get $x = 60 - 3y$, so the product is $P = (60 - 3y)y = -3y^2 + 60y$.

$y = -\frac{60}{2(-3)} = 10 \;\Rightarrow\; x = 60 - 30 = 30, \quad P = 30 \cdot 10 = 300$

The greatest product is $300$, at $x = 30$ and $y = 10$.

Let $x$ be each side out from the wall; the parallel side is $60 - 2x$ and the area is $A = x(60 - 2x)$, maximized at $x = -\tfrac{60}{2(-2)} = 15$.

$\text{parallel side} = 60 - 2(15) = 30$

The side parallel to the wall is $30$ metres (twice each of the other two). The value $450$ is the area, not a side.

Name the numbers $x$ and $x - 10$, so the product is $P = x(x - 10) = x^2 - 10x$. Since $a = 1 > 0$ it has a minimum at $x = -\tfrac{-10}{2} = 5$.

$P = (5)^2 - 10(5) = 25 - 50 = -25$

The numbers are $5$ and $-5$, and the least product is $-25$.

Expanded, $R = -2p^2 + 80p$, so the best price is $p = -\tfrac{80}{2(-2)} = 20$. Substitute back for the revenue.

$R = 20(80 - 2 \cdot 20) = 20 \cdot 40 = 800$

The greatest revenue is $800$ dollars, at a price of $20$ dollars selling $40$ items.

Since $a = 2 > 0$ the cost has a minimum at $x = -\tfrac{-24}{2(2)} = 6$. Substitute back.

$C = 2(6)^2 - 24(6) + 100 = 72 - 144 + 100 = 28$

The least cost is $28$ dollars, reached at $6$ units. The input $6$ is where it happens, not the cost.

With width $x$, the length is $100 - x$ (half the perimeter minus the width), so $A = x(100 - x) = 100x - x^2$. Since $a = -1 < 0$ it opens downward, so the maximum is at the vertex.

$x = -\frac{100}{2(-1)} = 50, \qquad A = 50(100 - 50) = 50 \cdot 50 = 2500$

The greatest area is $2500$ square metres, from a $50$ by $50$ square.

With one number $x$, the product $P = x(40 - x)$ is largest at the vertex $x = -\tfrac{40}{2(-1)} = 20$.

$x = 20 \;\Rightarrow\; \text{other} = 40 - 20 = 20$

The pair is $20$ and $20$. A fixed sum gives the largest product when it is split evenly.

Let $x$ be each side out from the barn; the parallel side is $240 - 2x$, so $A = x(240 - 2x) = -2x^2 + 240x$.

$x = -\frac{240}{2(-2)} = 60 \;\Rightarrow\; A = 60(240 - 120) = 60 \cdot 120 = 7200$

The largest area is $7200$ square feet, from a $60$ by $120$ pen.

The peak is at $t = -\tfrac{128}{2(-16)} = 4$ seconds. Substitute back for the height.

$h = -16(4)^2 + 128(4) = -256 + 512 = 256$

The maximum height is $256$ feet, reached at $t = 4$ seconds.