Question 1:
1 | (a) | DR
z = f(2, y) = 8 + 4y – 2y2
= 10 – 2(y – 1)2⇒ max at (1, 10) or (2, 1, 10)
Crossing z-axis at 8, y-axis at 1± 5and
showing (1,10) as a max | M1
A1
B1
A1
[4] | 1.1
1.1
1.1
1.1 | Deriving correct equation of graph
of section.
Finding TP by completing the
square, use of “–b/2a”,
differentiation or mid-point
between roots.
∩-shaped parabola which crosses
horizontal axis twice.
Coordinates of intercepts and max
must be shown on graph or
apparent in working. Allow
decimal values (awrt –1.2 and 3.2)
for the y-intercepts. | Working must be shown.
Condone incorrect variable names
on axes (eg x-y for y-z).
z intercept must be shown as
positive and max in 1st quadrant.
However, scale is unimportant
except that the negative y-intercept
must be closer to O than the
positive one.
(b) | ∂z
=3x2 +2xy
∂x
∂z
=x2 −4y
∂y
∂z
=0
∂x
⇒3x2 +2xy=0
2 3
⇒ either x=0 or x=− y or y=− x
3 2
∂z 2 4
=0, x=− y⇒4y= y2 ⇒ y=9
∂y 3 9
x = –6
z = –54 so (–6, 9, –54)
... or x=0⇒ y=0 so (0, 0, 0) | B1
B1
M1
M1
A1
A1
A1
[7] | 1.1
1.1
1.1
1.1
1.1
1.1
1.1 | Setting a partial derivative to 0
and deriving condition(s) on x
and/or y.
Substituting condition into other
partial derivative equation to
derive a non-zero value for x or y.
∂z 3
=0, y=− x
∂y 2
⇒x2 +6x=0⇒x=−6...
From correct working only.
∂z
Derived from both =0 and
∂x
∂z
=0 (could be by observation).
∂y | Or
∂z
=0
∂y
1
⇒ y= x2 or x2 =4y or x=±2 y
4
∂z 1
=0, y= x2
∂x 4
1
3x2 + x3 =0⇒x=−6
2
or
∂z
=0, x=±2 y
∂x
3
12y±4y2 =0⇒ y=9
y = 9
If an extra SP is presented then A1
can be awarded for either SP correct
and then A0.