Let the length of the frame be $l$ and the width be $w$. Then $2l+2w=60$, so $l+w=30$. Therefore, we want to maximize the area $lw$ subject to the constraint $l+w=30$. We can deduce some information by examining different possible widths and the resulting lengths.
If the width is 1, then the length is 29.
If the width is 2, then the length is 28.
If the width is 10, then the length is 20.
If the width is 20, then the length is 10.
If the width is 30, then the length is 0.
If we make the width any larger than 20, then the length will be negative. Therefore, of the choices given, the rectangle $\boxed{\textbf{(C) } 20 \text{ cm by } 10 \text{ cm}}$ gives the maximum area.
A rectangular frame must have a perimeter of 60 cm. What dimensions should the frame have to maximize the area within the frame?
A) 15 cm by 15 cm
B) 29 cm by 1 cm
C) 20 cm by 10 cm
D) 25 cm by 5 cm
3 answers
do it again
We are given that the perimeter of the rectangular frame is 60 cm. This can be expressed as $2l + 2w = 60$, where $l$ represents the length and $w$ represents the width.
We want to maximize the area within the frame, which is given by $A = lw$.
We can rewrite the equation $2l + 2w = 60$ as $l = 30 - w$ and substitute it into the area formula:
$A = (30 - w)w$
Expanding the equation, we have $A = 30w - w^2$.
Since the coefficient of $w^2$ is negative, the parabolic shape of this equation tells us that the area $A$ will be maximized when $w$ is at the vertex of the parabola.
The vertex of a quadratic equation in the form $ax^2 + bx + c$ can be found at $x = -\frac{b}{2a}$. In this case, $a = -1$ and $b = 30$, so the width $w$ that maximizes the area $A$ is $w = -\frac{30}{2(-1)} = 15$.
We can then substitute $w = 15$ back into the equation $l = 30 - w$ to find the length of the frame:
$l = 30 - 15 = 15$
Thus, the dimensions that maximize the area within the frame are $\boxed{\textbf{(A) } 15 \text{ cm by } 15 \text{ cm}}$.
We want to maximize the area within the frame, which is given by $A = lw$.
We can rewrite the equation $2l + 2w = 60$ as $l = 30 - w$ and substitute it into the area formula:
$A = (30 - w)w$
Expanding the equation, we have $A = 30w - w^2$.
Since the coefficient of $w^2$ is negative, the parabolic shape of this equation tells us that the area $A$ will be maximized when $w$ is at the vertex of the parabola.
The vertex of a quadratic equation in the form $ax^2 + bx + c$ can be found at $x = -\frac{b}{2a}$. In this case, $a = -1$ and $b = 30$, so the width $w$ that maximizes the area $A$ is $w = -\frac{30}{2(-1)} = 15$.
We can then substitute $w = 15$ back into the equation $l = 30 - w$ to find the length of the frame:
$l = 30 - 15 = 15$
Thus, the dimensions that maximize the area within the frame are $\boxed{\textbf{(A) } 15 \text{ cm by } 15 \text{ cm}}$.